matematika ii (mathematics ii)

Moderné vzdelávanie pre vedomostnú spoločnosťProjekt je spolufinancovaný zo zdrojov EÚ

MATEMATIKA II(MATHEMATICS II)

Stavebná fakulta

RNDr. Pavol Purcz, PhD.Doc.Ing. Roman Vodička, PhD.

NÁZOV: Matematika II (Mathematics II)AUTORI: Purcz Pavol, Vodička RomanVYDAVATEĽ: Technická univerzita v KošiciachROK: 2015NÁKLAD: 50 ksROZSAH: 171 stránISBN: 978-80-553-2046-5

Rukopis neprešiel jazykovou úpravou.Za odbornú a obsahovú stránku zodpovedajú autori.



Stavebná fakulta


Introduction

Aim

To provide theoretical knowledge required for studying specialized subjects and to apply obtained knowledge to the solution of technically orientedproblems basically using integral calculus and differential equations.

Contents and objectives

The aim of the module is to familiarize students with the necessary mathematical theory and its applications. At the beginning of each chapter aredefinitions of some terms and their properties required to solve the problems. As the publication is designed as a collection of problems, it neither containsdefinitions of all notions and terms nor proofs of mathematical theorems. The publication also includes solved examples and problems for solving andself-assessment according to the following content.

1. Function of several variables

2. Indefinite integral

3. Definite integral

4. Differential equations

5. Double integral

Prerequisities

Matematika I (Mathematics I. (P. Purcz, R. Vodička, TU Košice, 2012, ISBN 978-80-553-1279-8)

Introduction – 2



Stavebná fakulta


Táto publikácia vznikla za finančnej podpory z Európskeho sociálneho fondu v rámci Operačného programu VZDELÁVANIE.

Prioritná os 1 Reforma vzdelávania a odbornej prípravy.Opatrenie 1.2 Vysoké školy a výskum a vývoj ako motory rozvoja vedomostnej spoločnosti.

Názov projektu: Balík inovatívnych prvkov pre reformu vzdelávania na TUKE

NÁZOV: Matematika II (Mathematics II)AUTORI: Purcz Pavol, Vodička RomanVYDAVATEĽ: Technická univerzita v KošiciachROK: 2015NÁKLAD: 50 ksROZSAH: 171 stránISBN: 978-80-553-2046-5

Rukopis neprešiel jazykovou úpravou.Za odbornú a obsahovú stránku zodpovedajú autori.



Stavebná fakulta


Introduction

Aim

To provide theoretical knowledge required for studying specialized subjects and to apply obtained knowledge to the solution of technically orientedproblems basically using integral calculus and differential equations.

Contents and objectives

The aim of the module is to familiarize students with the necessary mathematical theory and its applications. At the beginning of each chapter aredefinitions of some terms and their properties required to solve the problems. As the publication is designed as a collection of problems, it neither containsdefinitions of all notions and terms nor proofs of mathematical theorems. The publication also includes solved examples and problems for solving andself-assessment according to the following content.

1. Function of several variables

2. Indefinite integral

3. Definite integral

4. Differential equations

5. Double integral

Prerequisities

Matematika I (Mathematics I. (P. Purcz, R. Vodička, TU Košice, 2012, ISBN 978-80-553-1279-8)

Introduction – 2

Chapter 1. (Function of several variables)

Mission

Show to students the real function of two or several real variables, generally, its properties, so-called partial derivative as a way to calculate derivativesin this case using calculus of one real functions of one real variable. Giving to students the basics of vector analysis, i.e. calculus of vector function,both, its properties and derivatives.

Objectives

1. Learning to determine the domain of function, mainly of two (or three variables respectively) focused on its geometric interpretation in the plane (orspace respectively).

2. Learning to calculate partial derivatives of functions of several variables, as well as the composed functions and higher order partial derivatives.

3. Being able to apply the knowledge obtained from the previous point to calculate the equation of the tangent plane and its normal line to the graphof the function of two variables as well as the local, and global extremes as well as extremes with respects to given sets.

4. Discover the concept of implicit function in the case of functions of one, or two variables, respectively, knowing calculate its derivative or partialderivatives, respectively and apply them to determine the equations of tangents and normals or tangent plane and normal line to the graph of afunction, respectively.

5. Understand concepts such as vector function, directional derivatives, gradients, divergence and rotation and apply them to know the role of vectoranalysis.

Prerequisite knowledge

differential calculus of the one function of the one real variables, analytical geometry, linear algebra

Chapter 1. (Function of several variables) – 1

Introduction

The subject of our further study will be a set of n- real numbers and real function, which can be defined on this set. Terms such as domain, boundedness,operations between functions, continuity or limits are equivalent concepts for real functions of one real variable and their definitions are very similar.Graph of the function of n variables is generally very difficult to see, because our imaging resources are quite limited. And therefore normally only chartshows the graph of the function of one or two variables. It is still possible to easily view domain of the function of three variables simultaneously and itis already everything that concerns our ability of the geometric visualization, since our existence in this world is given by the three-dimensional space.Therefore, the practical problems encountered in this chapter include (with few exceptions), just a function of two variables. The term derivative of thereal function of several variables is limited to the term so called partial derivative, and its calculation proceeds the same way as in the case of functionsof one variable.

Function of several variables – Domain of the function of several variables.

In determining the domain proceed similarly as in the function of one variable. Since the result is a set of points in the 2- or 3-dimensional space,respectively,it is suggested to reinforce the concepts of geometric sets and draw graphically formed situation.

A set of all n real numbers is called as n-dimensional Euclidean space En, if for each couple of the sets of n-real numbers A[a1, a2, ..., an] andB[b1, b2, ..., bn] is given following value:

%(A,B) =

√√√√ n∑i=1

(ai − bi)2,

which is called the difference (between the sets of n-real numbers).Let’s M ⊂ En. Then each mapping of M to one-dimensional Euclidean space is called a real function y of n variables. We denote its as y =f(x1, x2, ..., xn), or y = f(X). The set M is a domain of this function.

Excercise 1. Let’s find domain of the function

f(x, y, z) =1√

16− x2 − y2 − z2.

Solution.The domain is possible to determine from inequality 16− x2 − y2 − z2 > 0, or x2 + y2 + z2 < 16. the domain is created by the set of all points of X,which satisfy following condition %(O,X) < 4, what represents the interior of sphere with midpoint O[0, 0, 0] and radius equals 4 geometrically. �


Problems.

In problems 1 - 28 determine domain of the function several variables.1. z = ln(−x− y). 2. z = 1

y2−x2 .

3. z = 5√xy. 4. z =

√(4− x2 − y2)(x2 + y2 − 1).

5. u = ln(9−x2−y2−z2)√x2+y2+z2−4

. 6. z = arcsin x−1y.

7. z =√1− x2 +

√1− y2. 8. z = 1

25−x2−y2 .

9. z = arcsin(x+ y). 10. z =√y − x2 +

√1− y.

11. z =√x−√y. 12. z = ln(4−x2−y2)√

xy.

13. z =√

x2+y2−x2x−x2−y2 . 14. z = x2 + 2xy − 3y3.

15. z = 1x+ 1

y−1 . 16. z = x2y2x+|y| .

17. z =√3x− 6√

y. 18. z = 1√

x+|y|− 1

x−|y| .

19. z = y + arccosx. 20. z = ln(|x|+ y) + 1√y−x .

21. z = lnx2y√|y−x|

. 22. z = πy2√x2 − y2 + lnxy.

23. z =

√4x−y2

ln(1−x2−y2) . 24. z = ln(y2 − 4x+ 8).

25. z = ln(x−√y). 26. z = 1√x+y

+ 1√x−y .

27. u = arcsinx+ arcsin y + arcsin z. 28. z = arcsin[2y(1 + x2)− 1].

Function of several variables – Partial derivatives of the 1-st order of the function of several variables.

The term derivative of the real function of several variables is limited to a term so called partial derivatives, in computing which proceed in the same wayas in the case of functions of one variable. If we calculate derivatives with respect of one variable follow to fact that all other variables are consideredconstant.

Let’s given function f(x1, x2, ..., xn) defined in neighbourhood of the point A[a1, a2, ..., an]. If there exist a finite limit(∂f

∂xi

)A

= limxi→ai

f(a1, ..., ai−1, xi, ai+1, ..., an)− f(a1, ..., ai, ..., an)xi − ai

then its value is called the partial derivative of f(x1, x2, ..., xn) with respect to xi at point A It is denoted by:(∂f∂xi

)A, ∂f(A)

∂xi, f ′xi(A).


Excercise 2. Let’s find partial derivatives of the 1-st order of the function z = x3 + 7x2y5 + y2 − 2.

Solution.First, let’s calculate z′x. the variable y is considered constant now. It means, we calculate derivatives of z only with respect one variable x. Thus

z′x = 3x2 + 7.2xy5 = 3x2 + 14xy5.

Similarly we calculate z′y. The variable x we considered constant. We calculate again derivatives of z only with respect one variable y. Thus

z′y = 7x25y4 + 2y = 35x2y4 + 2y. �

Excercise 3. Let’s find partial derivatives of the 1-st order of the function z = e−yx at point A[1, 0].

Solution.The partial derivatives of the 1-st order are calculated as:

z′x = e−yxyx2,

z′y = e−yx

(− 1x

)= − 1

xe−

yx .

After substitution of the coordinates of the point A we have z′x(A) = 0, z′y(A) = −1. �

Problems.

In problems 29 - 61 calculate partial derivatives of the 1-st order of the given functions.29. z = x4 + y4 − 4x2y2. 30. z = x2y + y3

x4. 31. z = x

y2.

32. z = x√x2+y2

. 33. z = cosx2

y. 34. z = x sin(x+ y).

35. z = xy. 36. z = ln(x+ y2). 37. z = arctg yx.

38. u = (xy)z. 39. u = x

yz . 40. z = ln(x+ ln y).

41. z = 3x4y − 5xy2 + 2y. 42. z = x 3√y + y2√

x. 43. u = xy

z.

44. u = x3y2z − xy2z3. 45. z = (5x− y)4. 46. z = xy2 − yx+ 2√x.

47. z = xyy−x . 48. z = e−xy. 49. z = x

√x2 + y2.

50. z = ln(x− yex). 51. u = x2ey sin z. 52. z = (sinx)cos y.53. u = z sin(xyz)− y cos(yz) + sin(xz).

54. z = ln(x+√x2 + y2).

55. z = xyesin(πxy).


56. z = xeyx . 57. z = x3+y3

x2+y2. 58. z = arctg

√xy.

59. z = ln x−yx+y

. 60. u = (y tg z)lnx. 61. u = (2xy2 + z3)11.

62. Prove, that function z = f(x, y) satisfies the equation:a) xyz′x + x2z′y = 2yz, ak z = x2 sin(y2 − x2);b) xz′x + yz′y =

zln y

, ak z = exy ln y;

c) xz′x +y

lnxz′y = 2yz, ak z = xy;

d) yz′x − xz′y = 0, ak z = ln(x2 + y2);

e) 2xz′x + yz′y = 0, ak z = exy2 .

63. Prove, that function u = f(x, y, z) satisfies the equation:a) u′x

2 + u′y2 + u′z

2 = 1, ak u =√x2 + y2 + z2;

b) u′x + u′y + u′z = 1, ak u = x+ x−yy−z

Function of several variables – Total differential of function.

In the case of functions of two variables we can give partial derivatives at a given point some geometrical interpretation. There are tangent line at thispoint in the direction of the coordinate axes. Both of these tangent lines then generate the tangent plane to the surface to the graph of the function oftwo variables. Using analytical geometry we can easily find the equation of the normal line perpendicular to the plane at this point.

Let’s given the function of two variables z = f(X) = f(x, y), with its continual partial derivatives of the 1-st order at point A[a1, a2] . Then totaldifferential of the function f(X) at point A is defined as

df(A,X) =∂f(A)

∂x(x− a1) +

∂f(A)

∂y(y − a2),

or generally

dz =∂f

∂xdx+

∂f

∂ydy.

To calculate the approximate true relationship we usef(X) ≈ f(A) + df(A,X).

If the graph of the function z = f(x, y) is given and moreover f(x, y) is differentiable function, then the tangent plane to this graph at point M [x0, y0, z0]is given by following equation

∂f(M)

∂x(x− x0) +

∂f(M)

∂y(y − y0)− (z − z0) = 0.


The system of parametric equations of the normal line to this tangent plane at point M is

x = x0 +∂f(M)∂x

t,

y = y0 +∂f(M)∂y

t,

z = z0 − t.

Excercise 4. Let’s calculate total differential of function z = ln(x2 + y2).

Solution.First, let’s calculate the partial derivatives of the 1-st order. z′x =

2xx2+y2

a z′y =2y

x2+y2. Then

dz = 2xx2+y2

dx+ 2yx2+y2

dy. �

Excercise 5. Let’s calculate total differential of function z = xy2

at point A[1, 1].

Solution.First z′x =

1y2

and z′y = −2xy2

. After substitution z′x(A) = 1 and z′y(A) = −2. Then

df(A,X) = (x− 1)− 2(y − 1). �

Excercise 6. Let’s find approximate value of the expression (using total differential)√1, 023 + 1, 973.

Solution.Let’s consider function z =

√x3 + y3 and point A[1, 2]. Then it is possible calculate approximate value at point X[1, 02; 1, 97] using total differential√

1, 023 + 1, 973 ≈√13 + 23 +

(3x2

2√x3+y3

)A

(1, 02− 1) +

(3y2

2√x3+y3

)A

(1, 97− 2) =

= 3 + 0, 01− 2.0, 03 = 2, 95. �

Excercise 7. Let’s find the algebraic representation both, of the tangent plane and normal line to graph of the function z = x2+y2 at point M [1,−2, 5].

Solution.First, let’s calculate the partial derivatives of the 1-st order at point M . We have z′x(M) = 2, z′y(M) = −4. Then the algebraic representation of thetangent plan we can write as


2(x− 1)− 4(y + 2)− (z − 5) = 0,

or

2x− 4y − z − 5 = 0,

and the parametric representation of the normal line as

x = 1 + 2t,y = −2− 4t,z = 5− t. �

Problems.

In problems 64 - 69 find total differential of given functions.64. z = x2 − 2xy + y2. 65. z = ln

√x2 + y2. 66. z = ln cotg x

y.

67. z = ex ln y. 68. z = arctg x−y1+xy

. 69. u = sin(3x− 2y + 5z).In problems 70 - 78 find total differential of given functions at point A.

70. z = xyx2−y2 , A[3,−1]. 71. z = arcsin x

y, A[−1, 3].

72. u = z

√xy, A[1, 1, 1]. 73. z = x arctg(2y), A[−1, 1

2].

74. u = x2y + 2y2z + 3z2x, A[−1, 0, 1].75. z = x3 + 2xy2 − y3, A[2,−1], dx = −0, 1, dy = 0, 2.

76. z = ex2y, A[1, 1], dx = 0, 15, dy = −0, 05.

77. z = arctg yx, A[1, 2], dx = 0, 05, dy = 0, 01.

78. u = 2x sin y arctg z, A[−2, π2, 0], dx = 0, 03, dy = −0, 02, dz = 0, 04.

In problems 79 - 84 let’s find approximate value of the terms.

79.√

3, 032 + 4, 012. 80. 1, 052,01. 81. sin 151◦ cotg 41◦.

82.√

1, 023 + 1, 973. 83. 0, 971,05. 84. sin 29◦ tg 46◦.85. Find the approximate change both, in the diagonal and the surface area of a rectangle with sides x = 12 m, y = 9 m, if the first side is increased

by 2 cm and second is decreased by 4 cm.86. There are given height of the cone h = 15 cm and its radius of the base r = 8 cm. Find the approximate change of the volume of this cone, if

the height is decreased by 0, 3 cm a radius of the base is increased by 0, 2 cm.87. By deformation of the cylinder its radius is increased from 2 dm to 2, 05 dm, and its height is decreased from 10 dm to 9, 8 dm. Find the

approximate change of the volume of this cylinder.


88. Find the approximate change both, of the volume of the block and the area of its surface, if the lengths of its sides changes from 2 cm to 2, 04cm, from 3 cm to 2, 97 cm and from 4 cm to 4, 02 cm.

89. Find the approximate change of the volume of the cone, of the area of its surface and of the area of its surface casing, if its radius of the basechanges from 30 cm to 30, 1 cm and its height changes from 60 cm to 59, 5 cm.

90. Let’s find the algebraic representation both, of the tangent plane and normal line to graph of the function at given point.a) z = x2 + 4y2, T [3,−1, ?];b) z =

√x2 + y2 − xy, T [3, 4, ?];

c) z = x4 − 2x2y + xy + 2y, T [1, ?, 3];d) x+ y2 + z2 = 18, T [?, 3, 2];e) z = exy, T [1, 2, ?];f) z = ln(xy2) + x2y, T [1

4, 2, ?];

g) z = sin xy

, T [π, 1, ?];

h) z = 1xy

, T [1, 1, ?];

i) z = ln(x2 + y2), T [1, 0, ?].

Function of several variables – Partial derivative of the composed function.

Derivative of a composite function was in case of a function of one variable, a very important practical proposition. Even in the case of functions ofmore variables can it be decomposed into several components of the original function, which can be derivative separately. The structure functions canbe varied. We discuss now some of their variants.

A. If z = f(x, y) and x = ϕ(t) and y = ψ(t), then z = f (ϕ(t), ψ(t)) is a composed function of one variable t. If functions f , ϕ and ψ are differentiable,then is differentiable the composed function too and we have

dz

dt=∂z

∂x

dx

dt+∂z

∂y

dy

dt. (1.1)

Especially, if t = x, we getdz

dx=∂z

∂x+∂z

∂y

dy

dx. (1.2)

Excercise 8. Let’s calculate dzdt

of the function z = ex−2y, where x = sin t, y = t3.

Solution.Using (1.1) we have


dzdt

= ex−2y cos t− 2ex−2y3t2 = esin t−2t3(cos t− 6t2) . �

Excercise 9. Let’s calculate dzdx

of the function z = arctg(xy), where y = ex.

Solution.Using (1.2) we get

dzdx

= 11+x2y2

y + 11+x2y2

xex = ex+xex

1+x2e2x. �

B. If z = f(x, y) and x = ϕ(u, v) and y = ψ(u, v), then if functions f , ϕ and ψ are differentiable, then is differentiable the composed function too andwe have

∂z

∂u=∂z

∂x

∂x

∂u+∂z

∂y

∂y

∂u, (1.3)

∂z

∂v=∂z

∂x

∂x

∂v+∂z

∂y

∂y

∂v. (1.4)

Excercise 10. Let’s prove, that function z = arctg xy

, where x = u+ v, y = u− v satisfies the equation

∂z

∂u+∂z

∂v=

u− vv2 + u2

.

Solution.Using (1.3) and (1.4) we have

∂z∂u

= 1

1+x2

y2

1y+ 1

1+x2

y2

−xy2

= −x+yx2+y2

= −vu2+v2

,

∂z∂v

= 1

1+x2

y2

1y+ 1

1+x2

y2

xy2

= x+yx2+y2

= uu2+v2

,

thus

∂z∂u

+ ∂z∂v

= −vv2+u2

+ uv2+u2

= u−vv2+u2

.

Given functions satisfies the equation. �


Problems.

91. Let’s prove, that each differentiable function F (x, y) satisfies the equation:

a) y2F ′x − xyF ′y =y2

xF , ak F (x, y) = xf(x2 + y2)

b) yF ′x + xF ′y = 0, ak F (x, y) = f(x2 − y2)c) y2F ′x + xyF ′y = xF , ak F (x, y) = y

f(x2−y2)In problems 92 - 97 calculate dz

dt.

92. z = x2 + xy + y2, ak x = sin t, y = cos t.93. z =

√xy, ak x = sin t, y = t2.

94. z = exy ln(x+ y), ak x = t3, y = 1− t3.95. z = xln y, ak x = t3, y = t2.96. z = x

√y, ak x = ln t, y = 1 + et.

97. z = yx, ak x = et, y = 1− e2t.

In problems 98 - 103 calculate ∂z∂x

and dzdx

.98. z = arctg y

x, ak y = x2. 99. z = ln(ex + ey), ak y = x3.

100. z = arctg x+1y, ak y = e(x+1)2 . 101. z = xey, ak y = ϕ(x).

102. z = ln(x2 + y2), ak y = ϕ(x). 103. z = xy, ak y = ϕ(x).In problems 104 - 109 calculate ∂z

∂xand ∂z

∂y.

104. z = u+ v2, u = x2 + sin y, v = ln(x+ y).105. z = u2v − v2u, u = x cos y, v = x sin y.

106. z = uv, u = ln(x− y), v = exy .

107. z = u2 ln v, u = xy, v = 3x− 2y.

108. z = ueuv , u = x2 + y2, v = xy.

109. z = u2

v, u = x− 2y, v = y + 2x.

In problems 110 - 112 calculate partial derivatives of the 1-st order of the following functions.110. z = f(x2 − y2, exy). 111. z = f(xy + y

x). 112. z = f(x+ y, x− y).

Function of several variables – Higher order partial derivatives.

As with function of one variable, for its derivation we can look back an consider it as some other function, such as in the case of functions of severalvariables. The only one difference is in the fact that we have partial derivatives, which are independent of each other and hence opportunities to developother partial derivatives is suddenly much more. The number of combinations increases exponentially. Under certain theoretical assumptions, on theother hand, some combinations can be the same.


Let’s consider the function f(x1, x2, ..., xn) with its domain M . Let’s suppose, that this function has (in domain M1 ⊂ M) its partial derivative inrespect xi. f

′xi

is again a function of n-variables. If the new function has (in domain M2 ⊂ M1) partial derivative in respect xj, then this new partial

derivative is called as the 2-nd partial derivative of the origin function f(x1, x2, ..., xn) in respect xi and xj. This fact we denote as: ∂2f∂xi∂xj

or f ′′xixj . If

xi = xj, then ∂2f∂x2i

. There are exist account of the n2 partial derivatives of the 2-nd order, where n is a number of variables.

Theorem 1. If function f(x1, x2, ..., xn) has at point A[a1, a2, ..., an] differentiable partial derivatives f ′xi and f ′xj , then

f ′′xixj = f ′′xjxi .

We call this interchangeability of the partial derivatives of the 2-nd order.

Similarly is possible to define the partial derivatives of the 3-th, 4-th and next higher orders.

Excercise 11. Let’ calculate partial derivatives of the 2-nd order of the function z = x3 − 3x4y + y5 + 9.

Solution.First, partial derivatives of the 1-st order are f ′x = 3x2 − 12x3y, f ′y = −3x4 + 5y4.From last derivatives let’s calculate again next partial derivatives and we get:

f ′′xx = (3x2 − 12x3y)′x = 6x− 36x2y,f ′′xy = (3x2 − 12x3y)′y = −12x3,f ′′yx = (−3x4 + 5y4)′x = −12x3,f ′′yy = (−3x4 + 5y4)′y = 20y3.

Actually is possible to see, that f ′′xy = f ′′yx. �

Problems.

In problems 113 - 127 calculate partial derivatives of the 2-nd order of the given functions.113. z = xy + x

y. 114. z = xy.

115. z = y2

1+5x. 116. z =

√3xy + x2.

117. z = ln(s3 + t). 118. z = x3 − 3x4y + y5.119. z = 1

3xy. 120. z = e2y sinx.

121. z = xy + cos(x− y). 122. z = x sin(x+ y) + y cos(x+ y).123. z = arctg x−y

x+y. 124. z = exy.

125. z = cos2 yx. 126. z = x

√y + y

3√x .


127. z = arctg yx.

128. Let’s prove, that given differentiable functions satisfies following equation:a) z = ln(ex + ey), z′′xxz

′′yy − (z′′xy)

2 = 0b) z = arctg(2x− y), z′′xx + 2z′′xy = 0c) z = 2 cos2(x− y

2), 2z′′yy + z′′xy = 0

d) z = yexy , z′′xy − z′y + z′x = 0

e) z = ln√x2 + y2, z′′xx + z′′yy = 0

f) z = 2xy2 + cos(x+ y), z′′xy − z′′xx = 4y

Function of several variables – Extremes of the function of several variables.

Instead, the entire course of the investigation functions of several variables on the properties of the derivatives, and so on, we will now deal only graduallyfinding local extreme, global extremes and local extreme with respect to a set, i.e. maximum and minimum of the functions. Procedure for determiningthe local extremes is similar to the function of one variable and plays a key role here as the determination of stationary points. In determining theextremes with respect to a set enters into the calculation condition that limits the search to a given set of extremes. The concept of a global extremeis actually a maximum or minimum function on a closed set.

A. Let’s consider a function f(X) = f(x1, x2, ..., xn), with its continued partial derivatives of the 2-nd order. Let’s denote

aij = aji = f ′′xixj(A).

When searching for local extremes proceed as follows:

A. Calculate partial derivatives of the 1-st order f ′x1 , f′x2, ..., f ′xn .

B. Solve the system of equations f ′x1 = 0, f ′x2 = 0, ..., f ′xn = 0. Solutions are stationary points.

C. If at some stationary point A are values a11 > 0,

∣∣∣∣ a11 a12a21 a22

∣∣∣∣ > 0,

∣∣∣∣∣∣a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣ > 0, ... (it means, all sub-determinants are positive), then at

point A the function f has a local minimum.

If at some stationary point A are values a11 < 0,

∣∣∣∣ a11 a12a21 a22

∣∣∣∣ > 0,

∣∣∣∣∣∣a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣ < 0, ... (regularly sign changes), then at point A the function

f has a local maximum.


By irregular sign changes (for example

∣∣∣∣ a11 a12a21 a22

∣∣∣∣ < 0) at point A is no local extreme. In case det(A) equals 0 is impossible say any result.

Excercise 12. Find local extreme of the function z = x2 + y2 + xy − 6x− 9y.

Solution.Partial derivatives of the 1-st order must be zero.

z′x = 2x+ y − 6 = 0z′y = 2y + x− 9 = 0.

Solving previous system of equations we have x = 1, y = 4 These values are coordinates of stationary point A[1, 4] too.Next z′′xx = 2, z′′xy = z′′yx = 1, z′′yy = 2 and we have

a11 = 2 > 0,

∣∣∣∣ a11 a12a21 a22

∣∣∣∣ = ∣∣∣∣ 2 11 2

∣∣∣∣ = 3 > 0.

So it is clear, that at point A[1, 4] exist a local minimum of the given function and z(1, 4) = 21. �

B. Let’s consider function z = f(x, y) and let’s find its local extreme with respect to a set g(x, y) = 0. We proceed as follows:

A. If from equation g(x, y) = 0 is possible determine some variable, we calculate it and substitute to origin function. In next step we find localextreme of the function of one variable only.

B. If this process from previous point is impossible to do, we arrange so called Lagrange function

L(x, y) = f(x, y) + λg(x, y)

and we find local extreme of this new function. Stationary points with connected parameter λ is possible to determine from following equations:

L′x = 0, L′y = 0, L′λ = g(x, y) = 0.

Next steps as in previous case A..

Excercise 13. Find local extreme of the function z = 6− 4x− 3y with respect to a set x2 + y2 = 1.

Solution.Arrange Lagrange function and calculate its partial derivatives of the 1-st order


L(x, y, λ) = 6− 4x− 3y + λ(x2 + y2 − 1).L′x = −4 + 2λx = 0,L′y = −3 + 2λy = 0,L′λ = x2 + y2 − 1 = 0.

Solving last system of equations we have two stationary pointspre λA = 5

2, A[4

5, 35],

pre λB = −52, B[−4

5,−3

5].

Mme L′′xx = 2λ, L′′xy = 0, L′′yy = 2λ.

At stationary point A are a11 = 5 > 0,

∣∣∣∣ a11 a12a21 a22

∣∣∣∣ = ∣∣∣∣ 5 00 5

∣∣∣∣ = 25 > 0 so at point A exist local minimum of the origin problem.

At stationary point B are a11 = −5 < 0,

∣∣∣∣ a11 a12a21 a22

∣∣∣∣ = ∣∣∣∣ −5 00 −5

∣∣∣∣ = 25 > 0 so at point B exist local maximum of the origin problem. �

Problems.

In problems 129 - 151 determine local extremes of the following functions.129. z = 1 + 6y − y2 − xy − x2. 130. z = x2 + y2 − xy − x− y + 2.131. z = x2 + y2 + xy − 6x− 9y. 132. z = 4− (x− 2)2 − (y + 3)2.

133. u = 5 + 6z − 4z2 − 3t2. 134. z = 2x2 − 6xy + 5y2 − x+ 3y + 2.135. z = x2 − 2y2 − 3x+ 5y − 1. 136. z = x2 − y2 + 2x− 2y.137. z = x2 + (y − 1)2. 138. z = x3 + y3 − 18xy + 215.139. z = 27x2y + 14y3 − 69y − 54x. 140. z = e−x

2−y2(2y2 + x2).141. z = 5xy + 25

x+ 8

y, x > 0, y > 0. 142. z = x2 − (y − 1)2.

143. z = xy + x+ y − x2 − y2 + 2. 144. z = x2 + y2 − xy − 2x+ y.145. z = x3 − 3xy + y2 + y − 7. 146. z = 6xy − x3 − y3.147. z = e2x(x+ y2 + 2y). 148. z = e2x+3y(8x2 − 6xy + 3y2).149. z = xy + 50

x+ 20

y. 150. z = 2x3 + xy2 + 5x2 + y2.

151. z = 3x2y + y3 − 18x− 30y.In problems 152 - 159 determine local extremes of the following functions with respect to given sets.152. z = x2 + y2, 2x− y + 5 = 0. 153. z = xy, x+ y = 1.154. z = 1

x+ 1

y, 1x2

+ 1y2

= 14. 155. z = x+ 1

2y, x2 + y2 = 1.

156. z = x2 + 12xy + 2y2, 4x2 + y2 = 25. 157. z = x+ y, 1x2

+ 1y2

= 12.

158. z = 9− 8x− 6y, x2 + y2 = 25. 159. z = 8− 2x− 4y, x2 + 2y2 = 12.


In problems 160 - 169 determine global extremes of the following functions.160. z = x− 2y − 3, 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ x+ y ≤ 1.

161. z = x2 − 2y2 + 4xy − 6x− 1, x ≥ 0, y ≥ 0, y ≤ −x+ 3.162. z = x3 + y3 − 3xy, y ≥ −1, y ≤ 2, x ≥ 0, x ≤ 2.163. z = xy2(4− x− y), x = 0, y = 0, x+ y = 6.164. z = x2 + y2 − 12x+ 16y, x2 + y2 ≤ 25.165. z = x2 − xy + y2, |x|+ |y| ≤ 1.166. z = x− 2y − 1, x ≤ 0, y ≥ 0, y − x ≤ 1.167. z = x2 + 4xy + y2, |x|+ |y| ≤ 4.168. z = xy, x2 + y2 ≤ 2.169. z = x2 − y2, x2 + y2 ≤ 4.Next problems.170. In plane σ : x+ y− 2z = 0 find a point, the sum of the squares of the distances from the plane ρ1 : x+3z = 6 and ρ2 : y+3z = 2 is minimal.171. Among all of the blocks surface P choose one block with its maximal volume.172. Among all blocks with a solid diagonal length 2

√3 choose one with maximal volume.

173. Among all of the cylinder surface P = 6π choose one with maximal volume.174. Among all of the cone casing S choose one with maximal volume.175. Into hemisphere with a given radius R enter the block of the maximal volume.176. Into cone of its both, base radius R and height h enter the block of the maximal volume.177. Distribute a positive number a to three positive summands so, that their product will be maximal.178. In the plane locate such a point that the sum of the squares of the distances from the lines x = 0, y = 0 and x+ 2y − 16 = 0 is minimal.179. Into ellipsoid with half-axis a, b, c enter a block of the maximum volume so that its edges are parallel to the axes of the ellipsoid.180. Determine the dimensions of the concrete tank shape tetrahedral prism so that the concrete was minimal consumption for a given volume Vtank. Wall thickness are not considered.181. In plane x+ 2y − z + 3 = 0 determine a point, the sum of the squares of the distances from points [1, 1, 1] a [2, 2, 2] is minimal.

Function of several variables – Derivative of the function given by implicit form.

There are situations where it is not possible from the equation explicitly express one variable as a function of one or more variables, although we showthat the certain set can be such function defined. Then we talk about the function given by implicit form. With such expressed functions as a result ofany procedure we often meet, for example the analytical expression of the results of solving ordinary differential equations, which will be the subject ofstudy in one of the next chapter.


A. Let’s consider an equation F (x, y) = 0, with its point A[x0, y0]. If function F (x, y) is in the neighbourhood of the point A continuous and moreover

exist continuous partial derivatives up to 2-nd order, where ∂F (A)∂y6= 0, then using equation F (x, y) = 0 is determined an implicit function y = f(x) in

the neighbourhood of the point A. For derivatives of this function (first and second) holds:

y′ = −F′x

F ′y, (1.5)

y′′ = −Fxx + 2F ′′xyy

′ + F ′′yyy′2

F ′y(1.6)

Excercise 14. Let’s determine both derivatives, y′ and y′′ of the function y given implicitly by equation x2 − 2xy − y2 − 16 = 0 with point A[4, 0].

Solution.Calculate firstly both, partial derivatives of the 1-st and 2-nd order. We have:

F ′x = 2x− 2y, F ′y = −2x− 2y,F ′′xx = 2, F ′′xy = −2, F ′′yy = −2.

Partial derivatives are continuous across the all considered area and because F ′y(A) = −8 6= 0, using both, (1.5) and (1.6) we get:

y′ = − 2x−2y−2x−2y = x−y

x+y,

y′′ = −2−4x−yx+y−2(x−y

x+y )2

−2x−2y = 2y2+4xy−2x2(x+y)3

. �

B. Let’s consider an equation F (x1, x2, ..., xn, y) = 0, with its point A[a1, a2, ..., an, an+1]. If function F (x1, x2, ..., xn, y) is in the neighbourhood of the

point A continuous and moreover exist continuous partial derivatives of the 1-nd order, where ∂F (A)∂y6= 0, then using equation F (x1, x2, ..., xn, y) = 0 is

determined an implicit function y = f(x1, x2, ..., xn) in the neighbourhood of the point A. For its partial derivatives holds:

∂f

∂xi= −

F ′xiF ′y

. (1.7)

Excercise 15. Let’s determine derivatives of the 1-st order of the function z given implicitly by equation 4x2+2y2− 3z2+xy− yz+x− 4 with pointA[1, 1, 1].

Solution.Calculate firstly partial derivatives. We have:


F ′x = 8x+ y + 1, F ′y = 4y + x− z, F ′z = −6z − y.

Using (1.7) we get:

z′x = −8x+y+1−6z−y = 8x+y+1

6z+y,

z′y = −4y+x−z−6z−y = 4y+x−z

6z+y. �

C. Let’s consider an equation of the surface in implicit form F (x, y, z) = 0, then the equation, represent the tangent plane to this surface at pointM [x0, y0, z0] is possible write in form:

F ′x(M)(x− x0) + F ′y(M)(y − y0) + F ′z(M)(z − z0) = 0. (1.8)

Parametric representation of normal line to considered tangent line is possible write in form:

x = x0 + F ′x(M)t,y = y0 + F ′y(M)t,z = z0 + F ′z(M)t.

(1.9)

Excercise 16. Let’s determine an representation of the tangent plane and its normal line to graph of the function given implicitly x2− y2+ z2− 6 = 0at point A[1, 2,−3].

Solution.We have F ′x(A) = (2x)A = 2, F ′y(A) = (−2y)A = −4, F ′z(A) = (2z)A = −6.Using both, (1.8) and (1.9) we get following representation of the tangent plane

x− 2y − 3z − 6 = 0,

and Parametric representation of normal line

x = 1 + 2ty = 2− 4tz = −3− 6t. �


Problems.

In problems 182 - 185 calculate derivative y′ at point A of the function given implicitly by equation F (x, y) = 0.182. x2 − xy − y2 − 1 = 0, A[2, 1]. 183. 2y − 2x2 + y = 0, A[

√3, 2].

184. x sin y − y cosx = 0, A[14π,−1

4π]. 185. ln

√x2 + y2 − arctg y

x= 0, A[1, 0].

In problems 186 - 193 calculate derivative y′ of the function given implicitly by equation F (x, y) = 0.186. xy − ln(exy + e−xy) + ln 2 = 0. 187. ex cos y + ey cosx− 1 = 0.

188. xex − y2 − xy = 0. 189. x2 − 2xy − y2 − 16 = 0.190. x− ln y − y2 = 0. 191. x− y + 4 sin y = 0.

192. y3 − 4xy + x2 = 0. 193. xy − yx = 0.In problems 194 - 197 calculate partial derivatives z′x and z′y at point A of the function given implicitly by equation F (x, y, z) = 0.194. 3x2 − 4y2 + 2z2 − xy + xz − y = 0, A[1, 1, 1].195. cos2 x+ cos2 y + cos2 z − 2 = 0, A[1

3π, 1

2π, 1

6π].

196. z4 − 4xyz − 1 = 0, A[0, 2, 1].197. xez − yz − xy2 = 0, A[2, 1, 0].In problems 198 - 205 calculate partial derivatives z′x and z′y of the function given implicitly by equation F (x, y, z) = 0.198. 4x2 + 2y2 − 3z2 + xy − yz + x− 4 = 0. 199. x cos y + y cos z + z cosx− 1 = 0.

200. ez + x2y + z + 5 = 0. 201. x3 + 2y3 + z3 − 3xyz − 2y + 27 = 0.202. sinxy + sin yz + sin zx− 1 = 0. 203. z − xy sinxz = 0.204. z + ez − xy − 1− 0. 205. arctg x+ arctg y + arctg z − 5 = 0.206. Determine the equation of the tangent line and normal line at point T to the graph of the function y = f(x) given implicitly by equationF (x, y) = 0.

a) x5 + y5 − 2xy = 0, T [1, 1];b) x2 − y2 + 2x− 4y − 6 = 0, T [1,−1];c) 2x3 − x2y + 3x2 + 4xy − 5x− 3y + 6 = 0, T [0, 2];d) x3 + y3 − 2xy = 0, T [1, 1];

207. Determine the equation of the tangent plane and normal line to the graph of the function z = f(x, y) at point T given implicitly by equationF (x, y, z) = 0.

a) x2 − y2 + z2 − 6 = 0, T [1, 2,−3];b) 4 +

√x2 + y2 + z2 − x− y − z = 0, T [2, 3, 6];

c) x2 + y2 + z2 − 49 = 0, T [2,−6, ?];d) (z2 − x2)xyz − y5 − 5 = 0, T [1, 1, 2];


e) z − y − ln xz= 0, T [1, 1, 1];

f) x3 + y3 + z3 + xyz − 6 = 0, T [−1, ?, 1];g) ez − z + xy − 3 = 0, T [2, 1, ?];h) 8− 2

xz − 2

yz = 0, T [2, ?, 1].

Function of several variables – Basics of vector analysis.

The motivation for this part of the chapter is a number of physical applications, where we can find practical use of the following descriptions to describevarious physical phenomenons.

A. Scalar vector function. Let’s M is a set of real numbers. Then a function f , which assigned for each number t from M a vector f(t) is called thevector function of one real variable or scalar vector function. The set of M is called a domain of the function f and f(t) the value of the scalar vectorfunction f at value equals t.Let’s define in 3-D space rectangular coordinate system, then for vector function f holds:

f(t) = x(t)i+ y(t)j+ z(t)k,

orf(t) : x = x(t),

y = y(t),z = z(t),

where i, j, k are unit vectors. The function x(t) is called as 1-st component, the function y(t) as 2-nd component and the function z(t) as 3-thcomponent of the vector function f .At considered rectangular coordinate system f(t) = x(t)i+ y(t)j+ z(t)k has a function its derivative f at value equals t0 if and only if exist derivativesof the functions x(t), y(t) and z(t) at value equals t0 and holds:

f ′(t0) = x′(t0)i+ y′(t0)j+ z′(t0)k. (1.10)

Excercise 17. Let’s determine the equation of the tangent plane and normal line to the curve f(t) : x = t4, y = t−3t2, z = t−1 at point T [1,−2, 0].

Solution.First, let’s determine a value t0 corresponding to point T . Since the point T lies on the curve f(t), next holds:


f(t) : 1 = t4,−2 = t− 3t2,0 = t− 1.

Solving last equations we have t0 = 1. Because (1.10) holds, it possible to calculate derivative of the scalar function at t0 = 1.

f ′(1) = x′(1)i+ y′(1)j+ z′(1)k = (4t3)1i+ (1− 6t)1j+ 1k = 4i− 5j+ 1k.

This vector (4,−5, 1) is a directional vector of the tangent line and normal vector of the normal plane together. Parametric representation of tangentline is system:

x = 1 + 4ty = −2− 5tz = t.

And finally a normal plane to given curve is represents by equation:

4(x− 1)− 5(y + 2) + 1(z − 0) = 0,4x− 5y + z − 14 = 0. �

B. Directional derivative. Let’s consider a function of several variables f(X) = f(x1, x2, ..., xn), where X[x1, x2, ..., xn] with point A[a1, a2, ..., an]and unit vector l = (l1, l2, ..., ln). Then derivative of the function f(X) at point A in direction of vector l is called following limit

limt→0+

f(A+ tl)− f(A)t

and we denote this (definite) limit (if exist) as df(A)dl

.If f is a differentiable function of two variables f(x, y) at point A and vector l creates (in rectangular coordinate system) an angle α with coordinateox (in 2-D space), l = (cosα, sinα), then

df(A)

dl= f ′x(A) cosα + f ′y(A) sinα. (1.11)

If f is a differentiable function of three variables f(x, y, z) at point A and for vector l (in rectangular coordinate system) in 3-D space holds l =(cosα, cos β, cos γ), then

df(A)

dl= f ′x(A) cosα + f ′y(A) cos β + f ′z(A) cos γ. (1.12)

Excercise 18. Let’s calculate derivative of the function f(x, y, z) = xy2 + z3 − xyz at point A[1, 1, 2] in direction of vector l, which creates withcoordinates in rectangular system angles α = π

3, β = π

4, γ = π

3.


Solution.First, calculate partial derivatives of the function f at point A. We have

f ′x(A) = (y2 − yz)A = −1, f ′y(A) = (2xy − xz)A = 0, f ′z(A) = (3z2 − xy)A = 11.

Using term (1.12) we have

df(A)dl

= −1 cos π3+ 0 cos π

4+ 11 cos π

3= 5. �

C. Gradient. Let’s consider define a real function f(X) with its domain M , this situation is denoted as scalar field too. Let’s consider define a realfunction f(X) with its domain M , this situation is denoted as vector field.Let’s f(x) is a differentiable function at point A (in 3-D space). Then a gradient of the function f(X) at point A is called a vector grad f(A), whichsatisfies following conditions:

grad f(A) = f ′x(A)i+ f ′y(A)j+ f ′z(A)k. (1.13)

Let’s f is a differentiable function with its domain M . Then a gradient of the function f(X) (or gradient of the scalar field) f(X) is called a vectorfunction (with the same domain M), which satisfies following conditions:

grad f(X) = f ′x(X)i+ f ′y(X)j+ f ′z(X)k, (1.14)

for each point X ∈M . Let’s f(X) is a differentiable function at point A In 3-D (or 2-D) space, respectively. Then for derivative of the function f(X)at point A in direction determined by unit vector l holds:

df(A)

dl= grad f(A).l, (1.15)

df(A)dl

is a projection of the vector grad f(A) into vector l.

Excercise 19. Let’s determine a gradient of the function f(x, y, z) = x2 + y2 + z2 at point A[2,−2, 1].

Solution.First, calculate partial derivatives f at point A. We have

f ′x(A) = (2x)A = 4, f ′y(A) = (2y)A = −4, f ′z(A) = (2z)A = 2.

Using (1.13) we get

grad f(A) = 4i− 4j+ 2k. �


Excercise 20. Let’s determine the derivative of the function f(x, y, z) = 3x3− 4y3+2z4 at point A[2, 2, 1], in direction of vector l, if vector l = AB,where B[5, 4, 6].

Solution.From partial derivatives of the functions f at point A

f ′x(A) = (9x2)A = 36, f ′y(A) = (−12y2)A = −48, f ′z(A) = (8z3)A = 8.

is possible determine the gradient

grad f(A) = 36i− 48j+ 8k.

Next calculate coordinates of the vector l and its length

l = AB = B − A = (3, 2, 5), |l| =√9 + 4 + 25 =

√38.

Using (1.15) we get

df(A)dl

= (36i− 48j+ 8k) (3i+2j+5k)√38

= (36.3− 48.2 + 8.5) 1√38

= 52√38. �

D. Divergence a rotation. Let’s consider rectangular coordinate system of three vectors ordered in positive orientation and real function of threevariables:

f(X) = f1(X)i+ f2(X)j+ f3(X)k,

where functions f1(X), f2(X), f3(X), X[x, y, z] have their partial derivatives for set M in Euclidean space E3.A divergence of the vector function f(X) is called a vector function

div f =∂f1∂x

+∂f2∂y

+∂f3∂z

. (1.16)

A rotation of the vector function f(X) is called a vector function

rot f =

(∂f3∂y− ∂f2

∂z

)i+

(∂f1∂z− ∂f3∂x

)j+

(∂f2∂x− ∂f1

∂y

)k, (1.17)

This situation is possible write in form of one determinant symbolically so:

rot f =

∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂z

f1 f2 f3

∣∣∣∣∣∣ .Chapter 1. (Function of several variables) – 22

Excercise 21. Let’s determine both, divergence and rotation of the vector function f(X) = 5x2yi+ (x2 − z2)j +(2x− z2)k.

Solution.

div f = (5x2y)′x + (x2 − z2)′y + (2x− z2)′z = 10x+ 0− 2z.

rot f =((2x− z2)′y − (x2 − z2)′z

)i+ ((5x2y)′z − (2x− z2)′x) j+

((x2 − z2)′x − (5x2y)′y

)k =

= 2zi− 2j+ (2x− 5x2)k. �

Problems.

In problems 208 - 219 determine the equation of the tangent and normal plane to the curve.208. x = at, y = 1

2at2, z = 1

3at3, T [6a, 18a, 72a].

209. x = t, y = t2, z = t3, t0 = −1.210. x = sin 2t, y = 1− cos 2t, z =

√2 cos t, t0 =

π4.

211. x = t− sin t, y = 1− cos t, z = 4 sin t2, T [π

2− 1, 1, 2

√2].

212. x =√2 cos t, y =

√2 sin t, z = 4t, t0 =

π4.

213. x = t2, y = 1− t, z = t3, T [1, 0, 1].214. x = 3 tg t, y =

√2 cos t, z =

√2 sin t, t0 =

π4.

215. x = t3 − t2 − 5, y = 3t2 + 1, z = 2t3 − 16, t0 = 2.216. x = z, y = z2, T [−1, 1,−1].217. x2 + y2 + z2 = 50, x2 + y2 = z2, T [3,−4,−5].218. y2 + z2 = 25, x2 + y2 = 10, T [1, 3, 4].219. x2 + y2 + z2 = 3, x2 + y2 = 2, T [1, 1, 1].220. Find a point on the curve x = t3, y = t2 − 2t, z = 2t+ t2 at which the tangent is parallel to the plane 2x− 3y − 3z − 5 = 0.221. Find a point on the curve x = 2t2, y = 3t− t3, z = 3t+ t3 at which the tangent is parallel to the plane 3x+ y − z + 7 = 0.222. Find a tangent plane to graph of the ellipsoid x2 + 2y2 + z2 = 1, parallel to one another plane x− y + 2z = 0.223. Find a tangent plane to graph of the function x2 − y2 − 3z = 0 passing through point A[0, 0,−1] parallel to line x

2= y

1= z

2.

224. Find a tangent plane to graph of the ellipsoid 3x2 + 2y2 + z2 − 21 = 0 parallel to one another plane 6x+ 4y + z = 0.225. Find a tangent plane to graph of the function z = xy perpendicular to line x+2

2= y+2

1= z−1−1 .

In problems 226 - 231 calculate derivative in the direction of vector I given function at given point, if226. f(x, y) = 3x4 − x2y3 + y2, A[1,−1] and vector I create to rectangular coordinates so the angles α = π

6, β = π

3

227. f(x, y) = 3x2 − 6xy4 + 11y5, A[1, 1] and vector I is determined by vector B − A, where B[4, 5].228. f(x, y) = 5x2 − 6xy + 10y3 − 7, A[0, 1] and vector I = −i.229. f(x, y, z) = xy2 + z3 − xyz, A[1, 1, 2] and vector I creates with the coordinate axis angles α = π

3, β = π

4, γ = π

3.


230. f(x, y, z) = 5x2 − 2y2, A[2, 5, 0], B[−2, 2, 0] and vector I determined by vector B − A.231. f(x, y, z) = 3x3 − 4y3 + 2z4, A[2, 2, 1] and vector I is determined by vector B − A, where B[5, 4, 6].In problems 232 - 237 find the gradient of the given function.232. z = x2 − 6xy + y2 − 10x− 2y + 9. 233. z = x2 − 2xy + y2 + 4x− 8y − 7.234. z =

√x2 + y2. 235. u = xyzex+y+z.

236. u = arctg (x2 + 3y2 + z3). 237. u = x2 + 2y2 + 3z2 − xz + yz − xy.In problems 238 - 245 find the gradient of the given function at a given point.238. z = 2

x2+3y2, A[2,−1].

239. z = arctg xx+y

, A[1, 0].

240. u = x sin z − y cos z, A[2, 1, 0].241. z =

√4 + x2 + y2, A[1, 2].

242. u = x3 + y3 + z3 − 3xyz, A[2, 1,−1].243. u = x

x2+y2+z2, A[1, 2, 2].

244. z = x2 − 2xy + 3y2 + 4x− 2y + 5, A = [7, 4].245. u = x2 + 3y2 + 4z2 + xy − 2yz + 5xz, A[1, 0, 1].

246. Locate an angle between gradients of the function z = ln√x2 + y2 at points M [1, 1] and N [1,−1].

247. Locate an angle between gradients of the functions u = x3 + y3 + z3 a v = x2 − y2 + z2 at point M [1,−1, 1].248. Locate points in the plane in which the gradient of the function z = ln (x− 1

y) equals i+ j.

249. Locate an angle between gradients of the functions u = x2 + y2 − z2 and v = arcsin xx+y

at point A[1, 1,√7].

250. Locate points in the plane in which the gradient of the function u = ln 1+xyx

equals −169i+ j.

In problems 251 - 254 find the divergence of the vector function.251. u = x2i− xyj+ xyzk. 252. u = xy2z3(2i+ 3j− k).253. u = (yi+ zj+ xk) 1√

x2+y2+z2. 254. u = xyzi+ (2x+ 3y + z)j+ (x2 + z2)k.

In problems 255 - 258 find the divergence of the vector function at a given point.255. u = xyi+ (x2 − z2)j+ y

(x+z)k, A[1, 0, 2].

256. u = xyi+ y

zj+ z

xk, A[1, 1, 1].

257. u = exyzi+ sinx sin y sin zj+ ln (x+ y + z)k, A[0, 2,−1].258. u = (2x2y − 3xz3 + 5x3yz)i+ (4y3x+ xyz + 8z2)j+ (6z3xy2 − 7z2x+ 9zy)k, A[1, 1, 1].In problems 259 - 262 find the rotation of the vector function.259. u = x2i+ y2j+ z3k.260. u = xy2z3(2i+ 3j− k).261. u = (yi+ zj+ xk) 1√

x2+y2+z2.


262. u = (2x− 3y + 5z)i+ (x2 + 4y2 − 8z2)j+ (3x3 − y3 + 2z3)k.In problems 263 - 266 find the rotation of the vector function at a given point.263. u = xyi+ (x2 − z2)j+ y

(x+z)k, A[1, 0, 2].

264. u = xyi+ y

zj+ z

xk, A[1, 1, 1].

265. u = exyzi+ sinx sin y sin zj+ ln (x+ y + z)k, A[0, 2,−1].266. u = y2z2i+ x2z2j+ x2y2k, A[1, 2, 3].

Self-assessment questions

1. Which one equivalent exist at the theory of function of one variable equivalent to the term closed or open set, respectively, which is defined in thetheory of function of several variables?

2. It is possible at the theory of function of several variables to define terms equivalent to term one-side limit (from the right and left) used in the theoryof function of one variable?

3. Which conditions much be satisfied for equality to mixed partial derivatives of higher order?

4. Under what conditions is sufficient for the calculation of global extremes functions of several variables simultaneously consider cases only local extremes(or local extremes with respect to given sets)?

Conclusion

Since we live in three-dimensional space, all the events going on around us are physical point of view generally described functions of several variables.Therefore, it is important to study not only the real functions of one real variable, but several variables. Knowledge acquired in this chapter are necessaryfor studying other chapters of mathematical analysis, working with functions of several variables that have an irreplaceable role in describing and solvingphysics problems in engineering practice.

Bibliography


[1] Siagova, J.: Advanced Mathematics, Bratislava, STU, 2011.

[2] Bubenik, F.: Problems to Mathematics for Engineers, Praha, CVUT, 1999.

[3] Kreyszig, E.: Advanced Engineering Mathematics, New York 1993.

[4] Neustupa, J.: Mathematics I, Praha, CVUT, 1996.

[5] Neustupa, J.: Mathematics II, Praha, CVUT, 1998.

Solutions

Answers to self-assessment questions (if you are not able to formulate this answers), as well as many other answers can be found in the recommendedliterature.

1. half-plane x+ y < 0 2. all plane except for lines y = x, y = −x3. the interior both, of the first and third quadrant 4. 1 ≤ x2 + y2 ≤ 45. 4 < x2 + y2 + z2 < 9 6. |x− 1| ≤ |y|, y 6= 07. < −1, 1 > × < −1, 1 > 8. [x, y] : x2 + y2 6= 259. −1− x ≤ y ≤ 1− x 10. y ≥ x2, y ≤ 1

11. x ≥ 0, 0 ≤ y ≤ x2 12. x2 + y2 < 4, xy > 013. (x− 1

2 )2 + y2 ≥ 1

4 (x− 1)2+ y2 < 1 14. R3

15. all plane except for lines x = 0 and y = 1 16. all plane except for lines y = 2x, y = −2x, x < 017. x ≥ 0, y > 0 18. x > 0, |y| < x19. |x| ≤ 1, y ∈ R 20. y > x21. x < 0, y > 0; x > 0, y > x 22. x < 0, x ≤ y < 0;x > 0, 0 < y ≤ x23. part of plane inside parabola y2 = 4x between parabola and circle x2 + y2 = 1 except the top of parabola and circle points24. outside of the parabola y2 = 4(x− 2) 25. y < x2, y ≤ 026. y < x, y > −x 27. < −1, 1 >3

28. y ≥ 0, y ≤ 11+x2 29. z′x = 4x3 − 8xy2, z′x = 4y3 − 8x2y

30. z′x = 2xy − 4y3

x5 , z′y = x2 + 3y2

x431. z′x = 1

y2 , z′y = − 2x

y3

32. z′x = y2√(x2+y2)3

, z′y = − xy√(x2+y2)3

33. z′x = − 2x sin x2

y , z′y = − cosx2

y2

34. z′x = sin(x+ y) + x cos(x+ y), z′y = x cos(x+ y) 35. z′x = yxy−1, z′y = xy lnx

36. z′x = 1x+y2 , z

′y = 2x

x+y237. z′x = − y

x2+y2 , z′y = x

x2+y2

38. u′x = zx (

xy )z, u′y = − zy (

xy )z, u′z = (xy )

zln x

y 39. u′x = yuxz , u

′y = u ln x

z , u′z = −yuz2 lnx

40. z′x = 1x+ln y , z

′y = 1

y(x+ln y) 41. z′x = 12x3y − 5y2, z′y = 3x4 − 10xy + 2

42. z′x = 3√y − y2

2x√x, z′y = x

3 3√y2

+ 2y√x 43. u′x = yz

x u, u′y = zy(z−1)u lnx, u′z = yzu lnx ln y


44. u′x = 3x2y2z − y2z3, u′y = 2x3yz − 2xyz3, u′z = x3y2 − 3xy2z2 45. z′x = 20(5x− y)3, z′y = −4(5x− y)3

46. z′x = y2 = yx2 + 1√

x, z′y = 2xy − 1

x 47. z′x = y2

(y−x)2 , z′y = − x2

(y−x)2

48. z′x = −ye−xy, z′y = −xe−xy 49. z′x = 2x2+y2√x2+y2

, z′y = xy√x2+y2

50. z′x = 1−yexx−yex , z

′y = −ex

x−yex 51. u′x = 2xey sin z, u′y = x2ey sin z, u′z = x2ey cos z

52. z′x = cosx cos y(sinx)cos y−1

, z′y = − sin y ln sinx(sinx)cos y

53. u′x = yz2 cos(xyz) + z cos(xz), u′y = xz2 cos(xyz)− cos(yz) + yz sin(yz), u′z = sin(xyz) + xyz cos(xyz) + y2 sin(yz) + x cos(xz)

54. z′x = 1√x2+y2

, z′y = y

x2+y2+x√x2+y2

55. z′x = ( 1x + πy cos(πxy))z, z′y = ( 1y + πx cos(πxy))z

56. z′x = eyx (1− y

x ), z′y = e

yx 57. z′x = x4+3x2y2−2xy3

(x2+y2)2, z′y = y4+3x2y2−2x3y

(x2+y2)2

58. z′x = y√xy

2x(1+xy) , z′y = ln x

√xy

2(1+xy)59. z′x = 2y

x2−y2 , z′y = −2x

x2−y2

60. u′x = u ln(y tg z)x , u′y = u ln x

y , u′z =u ln x

cos z sin z 61. u′x = 22y2(2xy2 + z3)10, u′y = 44xy(2xy2 + z3)

10, u′z = 33z2(2xy2 + z3)

10

64. df = 2(x− y)dx+ 2(y − x)dy 65. df = xx2+y2 dx+ y

x2+y2 dy

66. df = −2(ydx− xdy)/y2 sin 2xy 67. df = ex(y ln ydx+ dy)/y

68. df = (1+y2)dx−(1+x2)dy1+x2+y2+x2y2

69. df = (3dx− 2dy + 5dz) cos(3x− 2y + 5z)

70. 532 (dx+ 3dy) 71. 1

12

√2(3dx+ dy) 72. dx− dy

73. π4 dx− dy 74. 3dx+ dy − 6dz 75. −3, 6

76. 14e 77. −0, 018 78. 0, 01

79. 9, 506 80. 1, 10 81. 0, 55582. 2, 95 83. 0, 97 84. 0, 502

85. −0, 8cm, −0, 3m2 86. 70, 37cm3 87. 3, 77dm3

88. 0, 36cm3; 0, 4cm2

89. −94, 25cm3; −16, 86cm2; 1, 99cm2

90. a) 6x− 8y − z − 13 = 0, x−36 = y+1

−8 = z−13−1 ;

b) 17x+ 11y + 5z − 60 = 0, x−317 = y−4

11 = z+75 ;

c) 2x− y + z − 3 = 0, x−12 = y−2

−1 = z−31 ;

d) x+ 6y + 4z − 31 = 0, x−51 = y−3

6 = z−24 ;

e) 2e2x+ e2y − z − 3e2 = 0, x−12e2 = y−2

e2 = z−e2−1 ;

f) 80x+ 17y − 16z − 52 = 0,x− 1

4

80 = y−217 =

z− 18

−16 ;

g) x− πy + z = 0, x−π−1 = y−1

π = z−1 ;

h) x+ y + z − 3 = 0, x−1−1 = y−1

−1 = z−1−1 ;

i) 2x− z − 2 = 0, x−12 = z

−1 , y = 0.

92. cos 2t 93. t cos t2√sin t

+√sin t

94. 0 95. 12t6 ln t−1 ln t96. 2+et(2+t ln t)

2t√1+et

97. −et − e−t


98. 11+x2 , − y

x2+y2 99. ex+3ex3x2

ex+ex3, ex

ex+ey

100. 1−2(x+1)2

y2+(x+1)2y, y

y2+(x+1)2101. ey + xeyϕ′(x), ey, kde y = ϕ(x)

102. 2x+2yϕ′(x)x2+y2 , 2x

x2+y2103. yxy−1 + xy lnxϕ′(x), yxy−1

104. z′x = 2x+ 2x+y ln(x+ y), z′y = cos y + 2

x+y ln(x+ y)

105. z′x = 3x2 sin y cos y(cos y − sin y), z′y = x3(sin y + cos y)(1− 3 sin y cos y)

106. z′x = vuv−1 1x−y + uv lnu 1

y exy , z′y = vuv−1 1

y−x + uv lnu−xy2 exy

107. z′x = 2xy2 ln(3x− 2y) + 3x2

y2 (3x− 2y), z′y = −2x2

y3 ln(3x− 2y)− 2x2

y2(3x−2y)

108. z′x = ex2+y2

xy x4−y4+2x3yx2y , z′x = e

x2+y2

xy y4−x4+2xy3

xy2

109. z′x = 2(x−2y)(x+3y)

(y+2x)2, z′y = (2x−y)(9x+2y)

(y+2x)2

110. z′x = 2xf ′u + yexyf ′v, z′y = −2yf ′u + xexyf ′v, u = x2 − y2, v = exy

111. z′x = y(1− 1x2 )f

′u, z

′y = (x+ 1

x )f′u, u = xy + y

x112. z′x = z′u + z′v, z

′y = z′u − z′v, u = x+ y, v = x− y

113. z′′xx = 0, z′′xy = 1− 1y2 , z

′′yy = 2x

y3

114. z′′xx = y(y − 1)xy−2, z′′xy = xy−1(1 + y lnx), z′′yy = xy ln2 x

115. z′′xx = 50y2

(1+5x)3, z′′xy = −10y

(1+5x)2, z′′yy = 2

1+5x

116. z′′xx = −9y2

4(3xy+x2)32, z′′xy = 9xy

4(3xy+x2)32, z′′yy = −9x2

4(3xy+x2)32

117. z′′ss =3s(2t−s3)(s3+t)2

, z′′st =−3s2

(s3+t)2, z′′tt =

−1(s3+t)2

118. z′′xx = 6x− 36x2y, z′′xy = −12x3, z′′yy = 20y3

119. z′′xx = 23x2y , z

′′xy = 1

3x2y2 , z′′yy = 2

3xy3

120. z′′xx = −e2y sinx, z′′xy = 2e2y cosx, z′′yy = 4e2y sinx121. z′′xx = −cos(x− y), z′′xy = 1 + cos(x− y), z′′yy = − cos(x− y)122. z′′xx = (2− y) cos(x+ y)− x sin(x+ y), z′′xy = (1− y) cos(x+ y)− (1 + x) sin(x+ y), z′′yy = −(2 + x) sin(x+ y)− y cos(x+ y)

123. z′′xx = −2xyv2 , z′′xy = x2−y2

v2 , z′′yy = 2xyv2 , v = x2 + y2

124. z′′xx = y2exy, z′′xy = (1 + xy)exy, z′′yy = x2exy

125. z′′xx = 2 cos2 yx3 , z′′yy = −2 cos 2y

x , z′′xy = sin 2yx2

126. z′′xx = 4y

9x73, z′′yy = −x

4√y3, z′′xy = 1

2√y −

1

3x43

127. z′′xx = 2xy(x2+y2)2

, z′′xy = y2−x2

(x2+y2)2, z′′yy = −2xy

(x2+y2)2

129. max = z(−2, 4) = 13 130. min = z(1, 1) = 1 131. min = z(1, 4) = −21132. max = z(2,−3) = 4 133. max = u( 34 , 0) =

294 134. max = z(−2,− 3

2 ) =34

135. ∅ 136. ∅ 137. max = z(0, 1) = 0138. min = z(6, 6) = −1139. min = z(1, 1) = −82, max = z(−1,−1) = 82140. min = z(0, 0) = 0, max = z(0, 1) = z(0,−1) = 2

e


141. min = z( 52 ,45 ) = 30 142. ∅ 143. max = z(1, 1) = 3

144. max = z(1, 0) = −1 145. min = z(1, 1) = −7 146. max = z(2, 2) = 8147. min = z( 12 ,−1) = −

e2 148. min = z(0, 0) = 0 149. max = z(5, 2) = 30

150. min = z(0, 0) = 0, max = z(− 53 , 0) =

12527

151. min = z(1, 3) = −72, max = z(−1,−3) = 72152. min = z(−2, 1) = 5 153. max = z( 12 ,

12 ) =

14

154. max = z(2√2, 2√2) =

√22 ,min = z(−2

√2,−2

√2) = −

√22

155. max = z( 25√5, 15√5) = 1

2

√5, min = z(− 2

5

√5,− 1

5

√5) = − 1

2

√5

156. max = z( 32 , 4) = z(− 32 ,−4)

4254 , min = z(2,−3) = z(−2, 3) = −50

157. min = z(2, 2) = 4, max = z(−2,−2) = −4158. max = z(−4,−3) = 59, min = z(4, 3) = −41159. min = z(2, 2) = −4, max = z(−2,−2) = 20160. min z = −5, max z = −2161. min = z(0, 3) = −19, max = z(0, 0) = −1162. max = z(2,−1) = 13, min = z(0,−1) = z(1, 1) = −1163. max = z(1, 2) = 4, min = z(2, 4) = −64164. min z = −75, max z = 125165. min z = 0, max z = 1166. max = z(0, 0) = −1, min = z(0, 1) = −3167. max = z(2, 2) = z(−2,−2) = 24, min = z(2,−2) = z(−2, 2) = −8168. max = z(1, 1) = z(−1,−1) = 1, min = z(1,−1) = z(−1, 1) = −1169. max = z(2, 0) = z(−2, 0) = 4, min = z(0,−2) = z(0, 2) = −4170. (3,−1, 1) 171.

√P6

172. 2, 2, 2

173. 1, 2 174.√

S√3π,√

2S√3π

175. 23

√3R, 2

3

√3R, 1

3

√3R

176. 23

√2R, 2

3

√2R, 1

3h 177. 13a,

13a,

13a 178. 8

5 ,165

179. 2a√3, 2b√

3, 2c√

3 180. 3√2V , 3

√2V , 1

23√2V 181. ( 12 ,−

12 ,

52 )

182. 34 183. 4

√3

1+4 ln 2184. π+4

π−4

185. 1 186. y′ = − yx 187. y′ = ey sin x−ex cos yey cos x−ex sin y

188. y′ = ex+xex−y2y+x

189. y′ = x−yx+y

190. y′ = y1+2y2

191. y′ = 11−4 cos y 192. y′ = 4y−2x

3y2−4x 193. y′ = y2

x21−ln x1−ln y

194. − 65 , 2 195. −1, 0 196. 2, 0

197. 0, 4 198. z′x = 8x+y+16z+y , z′y = x+4y−z

6z+y

199. z′x = z sin x−cos ycos x−y sin z , z

′y = x sin y−cos z

cos x−y sin z 200. z′x = − 2xyez+1 , z

′y = − x2

ez+1

201. z′x = x2−yzxy−z2 , z

′y = 6y2−3xz−2

3(xy−z2) 202. z′x = y cos xy+z cos xzx cos xz+y cos yz , z

′y = −x cos xy+z cos yz

x cos xz+y cos yz

203. z′x = y sin xz+xyz cos xz1−x2y cos xz , z′y = x sin xz

1−x2y cos xz204. z′x = y

ez+1 , z′y = x

ez+1


205. z′x = − 1+z2

1+x2 , z′y = − 1+z2

1+y2

206. a) x+ y − 2 = 0, x− y = 0;b) 2x− y − 3 = 0, x+ 2y + 1 = 0;c) x+ y − 2 = 0, x− y = 0;d) x− y = 0, x+ y − 2 = 0.

207. a) x− 2y − 3z − 6 = 0, x−11 = y−2

−2 = z+3−3 ;

b) 5x+ 4y + z − 28 = 0, x−25 = y−3

4 = z−61 ;

c) 2x− 6y + 3z − 49 = 0, x−24 = y+6

−12 = z−36 , 2x− 6y − 3z − 49 = 0, x−2

4 = y+6−12 = z+3

−6 ;

d) 2x+ y + 11z − 25 = 0, x−12 = y−1

1 = z−211 ;

e) x+ y − 2z = 0, x−11 = y−1

1 = z−1−2 ;

f) 5x+ 11y + z − 18 = 0, x+15 = y−2

11 = z−11 ;

g) x+ 2y − 4 = 0, x−21 = y−1

2 , z = 0;

h) x+ y − 4z = 0, x−21 = y−2

1 = z−1−4 .

208. x+ 6y + 36z − 2706a = 0, x−6a1 = y−18a

6 = z−72a36

209. x− 2y + 3z + 6 = 0, x+11 = y−1

−2 = z+13

210. 2y − z − 1 = 0, y−12 = z−1

−1 , x− 1 = 0

211. x+ y +√2z − π

2 − 4 = 0,x−π2 +1

1 = y−11 = z−2

√2√

2

212. x− y − 4z + 4π = 0, x−1−1 = y−1

1 = z−π4

213. 2x− y + 3z − 5 = 0, x−12 = y

−1 = z−13

214. 6x− y + z − 18 = 0, x−36 = y−1

−1 = z−11 215. 2x+ 3y + 6z − 37 = 0, x+1

2 = y−133 = z

6

216. x− 2y + z + 4 = 0, x+11 = y−1

−2 = z+11 217. 4x+ 3y = 0, x−3

4 = y+43 , z + 5 = 0

218. 12x− 4y + 3z − 12 = 0, x−112 = y−3

−4 = z−43 219. x− y = 0, x−1

1 = y−1−1 , z − 1 = 0

220. (0, 0, 0), (8, 0, 8) 221. (0, 0, 0), (8,−2, 14)222. x− y + 2z ±

√112 = 0 223. 4x− 2y − 3z − 3 = 0

224. 6x+ 4y + z ± 21 = 0 225. 2x+ y − z − 2 = 0

226. 7√3− 2, 5 227. 24, 8

228. 6 229. 5230. −4 231. 52√

38232. (2x− 6y − 10)i+ (2y − 6x− 2)j 233. (2x− 2y + 4)i+ (2y − 2x− 8)j234. 1√

x2+y2(xi+ yj) 235. ex+y+z[(1 + yz)i+ (1 + xz)j+ (1 + xy)k]

236. 11+(x2+3y2+z3)2

(2xi+ 6yj+ 3z2k) 237. (2x− y − z)i+ (4y − x+ z)j+ (6z − x+ y)k

238. 449 (−2, 3) 239. (0,− 1

2 )240. (0,−1, 2) 241. 1

3 i+23 j

242. 15i+ 9j− 3k 243. 7i−4j−4k81

244. 10i+ 8j 245. 7i− j+ 13k246. ϕ = π

2 247. ϕ = 0248. (0, 1), (2, 1)) 249. ϕ = π

2


250. ( 34 ,−13 ), (−

34 ,

73 ) 251. x(1 + y)

252. 2y2z3 + 6xyz3 − 3xy2z2 253. − 1√(x2+y2+z2)3

(xy + yz + xz)

254. z(y + 2) + 3 255. 0256. 3 257. −1258. 42 259. 0260. (−2xyz3 − 9xy2z2)i+ (6xy2z2 − y2z3)j+ (3y2z3 − 4xyz3)k261. − 1√

(x2+y2+z2)3[(x2 + y2 + xy)i+ (y2 + z2 + zy)j+ (x2 + z2 + xz)k]

262. (−3y2 + 16z)i+ (5− 9x2)j+ (2x− 3)k263. − 11

3 i− k 264. i+ 2j+ k265. i+ j− (sin 2 sin 1)k 266. −2i+ 16j− 18k


Chapter 1. (Indefinite integral)

Mission

Explain students concepts as antiderivative and indefinite integral, the fundamental formulas for the integration of real function of one real variable andteach students various methods and techniques of integration.

Objectives

1. Understand the concepts of antiderivative and indefinite integral, learn the basic formula for integrating and be able to apply them to solve simpleproblems.

2. Learning to use both, the substitution method and method by parts as two basic methods used for finding any antiderivative.

3. Being able to integrate all types of so called partial fractions and then manage the integration of rational functions of fraction decomposition methodinto partial fractions.

4. Learn the methods of integrating certain types functions containing square root (or higher) and finally, trigonometric functions.

Prerequisite knowledge

differential calculus of the function of one real variable, linear algebra, algebraic equations

Chapter 1. (Indefinite integral) – 1

Introduction

In this chapter we first introduce the notion of antiderivative and indefinite integral and basic formulas and rules for calculating the antiderivative orindefinite integral, respectively. Then we will discuss next methods for finding the antiderivative for some of the more complex functions. This process isalso called as integration process. It is the reverse task to finding the derivative of the function. Integration is much more difficult to perform than theprocess of differentiation and there are functions that can not be integrated naturally. Mastery learning content of this chapter is the key to all otherchapters that follow behind her. The problem of finding antiderivatives to the function is a leader for mostly of physician problems. E.g. if we fix somecoordinates of a moving point, the speed we know, among other things, we must establish a primitive function to function, which indicates the speedat each time point. With many other problems where it is necessary to determine the antiderivative we will meet in next three chapters.

Indefinite integral – Antiderivative, indefinite integral, basic rules

Basic rules for the process of integration is possible to get simply using the base rules for process of differentiation.

A. We say that a functionF (x) on the interval (a, b) is an antiderivative to the function f(x), if for each x ∈ (a, b) holds that

F ′(x) = f(x).

Each function F (x) + C, where C is any real number is an antiderivative to f(x) too.B. The set of all antiderivatives to f(x) is called as indefinite integral of the function f(x) which is denoted as

∫f(x) dx. Thus∫

f(x) dx = F (x) + C,

where C is any real number, which is denoted as constant of integration.C. Basic rules

a)∫xα dx = xα+1

α+1+ C (α real, α 6= −1),

b)∫

1xdx = ln |x|+ C,

c)∫ax dx = ax

ln a+ C,

d)∫ex dx = ex + C,

e)∫sinx dx = − cosx+ C,


f)∫cosx dx = sinx+ C,

g)∫

1cos2 x

dx = tg x+ C,

h)∫

1sin2 x

dx = − cotg x+ C,

i)∫

1√a2−x2 dx = arcsin x

a+ C,

j)∫

1a2+x2

dx = 1aarctg x

a+ C,

k)∫

1√x2+a

dx = ln |x+√x2 + a|+ C,

l)∫ f ′(x)

f(x)dx = ln |f(x)|+ C.

Basic properties of the indefinite integral ∫(f(x)± g(x)) dx =

∫f(x) dx±

∫g(x) dx,∫

kf(x) dx = k

∫f(x) dx, (k is a constant).

Excercise 1. Let’s evaluate following indefinite integrals

a)

∫cos 2x

cos2 x sin2 xdx, b)

∫3x√x− x6 + x3 cosx

x3dx, c)

∫1√

4− 4x2dx.

Solution.

a)

∫cos 2x

cos2 x sin2 xdx =

∫cos2 x− sin2 x

cos2 x sin2 xdx =

∫cos2 x

cos2 x sin2 xdx−

∫sin2 x

cos2 x sin2 xdx =

=

∫1

sin2 xdx−

∫1

cos2 xdx = − cotg x− tg x+ C,

b)

∫3x√x− x6 + x3 cosx

x3dx =

∫3x√x

x3dx−

∫x6

x3dx+

∫x3 cosx

x3dx =

= 3

∫x−

32 dx−

∫x3 dx+

∫cosx dx = − 6√

x− x4

4+ sinx+ C,


c)

∫1√

4− 4x2dx =

∫1√

4(1− x2)dx =

1

2

∫1√

1− x2dx =

1

2arcsinx+ C. �

Problems

Using basic rules determine following indefinite integrals.1.

∫ (3x2 + 2x+ 1 + 1

3x

)dx. 2.

∫11x9+12

x4dx.

3.∫ (1+

√x)2√x

dx. 4.∫ (

1√x− 1

4√x3

)dx.

5.∫ √

x4+4x2+4x3

dx. 6.∫2x(x2 − 5 + 3

x2) dx.

7.∫ex(1− e−x

x2

)dx. 8.

∫ax(1− a−x

x4

)dx.

9.∫2x 5x dx. 10.

∫(2x + 3x)2 dx.

11.∫

3.2x−2.8x6x

dx. 12.∫ (3x−4x)2

12xdx.

13.∫ (4x−5x)2

20xdx. 14.

∫ (5 cosx−

√3x5 + 3

1+x2

)dx.

15.∫tg2 x dx. 16.

∫cotg2 x dx.

17.∫

cos 2xcos2 x sin2 x

dx. 18.∫

1cos2 x sin2 x

dx.19.

∫sin2 x

2dx. 20.

∫cos2 x

2dx.

21.∫

cos 2xcosx−sinx dx. 22.

∫1+cos2 x1+cos 2x

dx.

23.∫

x2

1+x2dx. 24.

∫x2−1x2+1

dx.

25.∫

4−x21+x2

dx. 26.∫

x4

1+x2dx.

27.∫

x5−16x+2x2+4

dx. 28.∫ (1−x)2

x2dx.

Using rule No.12 determine next indefinite integrals.29.

∫x

x2−3 dx. 30.∫

x2

x3+1dx.

31.∫

2x−53x2−15x+22

dx. 32.∫(tg x+ cotg x) dx.

33.∫

sin 2x2 cos2 x+3

dx. 34.∫

e2x

1−3e2x dx.

35.∫

1x lnx

dx. 36.∫

1√1−x2 arcsinx dx.


Indefinite integral – Integration by substitution

Rarely is satisfactory to use the basic formulas for integration only. Even small deviations that change sign, and the constant into the formulas of thefunctions leads to the necessity of introducing substitution. This method is very important and very often used in the integration process. Accurateapplication and use of substitution in indefinite integral are described by following two basic rules.

A. The 1st rule of substitution ϕ(x) = z.An indefinite integral ∫

f(ϕ(x))ϕ′(x) dx (1)

is evaluated formally using a new variable z by substitution ϕ(x) = z and ϕ′(x) dx = dz and replacing into origin problem. We get∫f(ϕ(x))ϕ′(x) dx =

∫f(z) dz.

Finally we evaluated the indefinite integral on the right hand side and replace again z = ϕ(x).

Excercise 2. Let’s evaluate following indefinite integrals

a)

∫ √arctg x

1 + x2dx, b)

∫cosx

3 + sin xdx.

Solution.

a) The problem is according to 1st rule, where ϕ(x) = arctg x. So we use the substitution arctg x = z; 11+x2

dx = dz and we get∫ √arctg x

1 + x2dx =

∫ √z dz =

z32

32

+ C =2

3

√arctg3 x+ C.

b) The nominator is a derivatives of the denominator. Using last basic rule (No.12) we have∫cosx

3 + sin xdx = ln |3 + sin x|+ C. �


B. The 2nd rule of substitution x = ϕ(z).An indefinite integral ∫

f(x) dx

is evaluated formally using substitution x = ϕ(z) and dx = ϕ′(z)dz. After replacing into origin problem we have∫f(x) dx =

∫f(ϕ(z))ϕ′(z) dz.

We evaluate the indefinite integral on the right hand side and finally is necessary express z using x, z = g(x) and put retrospectively.

Excercise 3. Let’s evaluate indefinite integral ∫1

x2√1 + x2

dx.

Solution. Using substitution x = 1t

and dx = − 1t2dt we get

∫dx

x2√1 + x2

=

∫ − 1t2dt

1t2

√1 + 1

t2

= −∫

dt√t2+1t2

= −∫

t dt√t2 + 1

= −√t2 + 1 + C =

= −√1 + x2

x+ C. �

Problems

Using methods of substitution determine following indefinite integrals.37.

∫(x+ 2)9 dx. 38.

∫ √x+ 3 dx.

39.∫

15−x dx. 40.

∫5

6x−1 dx.

41.∫

2(x+3)2

dx. 42.∫

2(2x+3)4

dx.

43.∫x(x2 − 4)11 dx. 44.

∫sin(2x) dx.

45.∫cos(x

3) dx. 46.

∫tg(4x− 1) dx.

47.∫

19x2+1

dx. 48.∫

1√1−4x2 dx.

49.∫e−x dx. 50.

∫e2x dx.

51.∫ √

ex dx. 52.∫x2ex

3dx.


53.∫

x2

x6+4dx. 54.

∫x2

(1−x)3 dx.

55.∫

2x2

cos2(x3+1)dx. 56.

∫sin2 x cosx dx.

57.∫

sinx4√cosx dx. 58.

∫lnx−2x(lnx)2

dx.

59.∫

1+x√1−x2 dx. 60.

∫1−2 sinxcos2 x

dx.

61.∫

arcsinx−x√1−x2 dx. 62.

∫tg x

ln3(cosx)dx.

63.∫

e1/x

x2dx. 64.

∫e√x

√xdx.

65.∫

2x√1−4x dx. 66.

∫1

cos2 x√tg x−1 dx.

Indefinite integral – Integration by parts

The method of integration by parts we obtain easily from the formula for differentiation of the product of two functions. Using this method is possiblefind an antiderivatve of the some functions in a special format. Principle of the method is contained in the following theorem.

If functions u and v of the one real variable x have continuous derivatives u′ and v′ on interval (a, b) then following formula holds on this interval:∫uv′ dx = uv −

∫u′v dx. (2)

The rule (2) is possible to use for finding the indefinite integral in some cases of product of two functions. One of them we choose as u and the otheras v′. The choice needs to be done so that we know to calculate v by integration of v′ and moreover the new problem

∫u′v dx will be simpler than

the origin one. It’s dependent on own experience, which function is better choose as u and v′, respectively. However, in some cases, we can see someprinciples. If P (x) is a polynomial, then, for example by computing

∫P (x) arcsinx dx we put u = arcsinx, v′ = P (x). We can make the choice of

the replacing similarly, if we replace arcsinx by any another cyclometric function or logarithmic one. On the other side, for example, by computing∫P (x) sinx dx we choose u = P (x), v′ = sinx. We can make the choice of the replacing similarly, if we replace sinx by any another goniometric

function or exponential one.Sometimes is necessary the formula (2) to reuse several times. It may be that we get back to the original integral. In this case, we have the integralequation, from which is possible express the finding result.

Excercise 4. Let’s calculate following indefinite integrals

a)

∫(2x− 1)ex dx, b)

∫(x2 + 3x− 3) cosx dx, c)

∫ln(2x+ 3) dx, d)

∫ex cosx dx.

Solution.


a) We replace u = 2x− 1, v′ = ex. Then u′ = 2, v =∫ex dx = ex. We get∫

(x− 1)ex dx = (x− 1)ex −∫

2ex dx = (x− 1)ex − 2ex + C = (x− 3)ex + C.

b) We replace u = x2 + 3x− 3, v′ = cosx. Then u′ = 2x+ 3, v = sinx and∫(x2 + 3x− 3) cosx dx = (x2 + 3x− 3) sinx−

∫(2x+ 3) sinx dx.

This method we use again the second times. Now, we replace u = 2x+ 3, v′ = sinx. Then u′ = 2, v = − cosx and∫(x2 + 3x− 3) cosx dx = (x2 + 3x− 3) sinx−

[(2x+ 3)(− cosx)−

∫−2 cosx dx

]=

= (x2 + 3x− 3) sinx+ (2x+ 3) cosx− 2 sinx+ C.

c) We replace u = ln(2x+ 3), v′ = 1. Then u′ = 22x+3

, v = x and∫ln(2x+ 3) dx = x ln(2x+ 3)−

∫2x

2x+ 3dx.

∫2x

2x+ 3dx =

∫2x+ 3− 3

2x+ 3dx =

∫ (1− 3

2x+ 3

)dx = x− 3

∫1

2x+ 3dx =

= x− 3

2

∫2

2x+ 3dx = x− 3

2ln |2x+ 3|+ C.

After replacing we have ∫ln(2x+ 3) dx = x ln(2x+ 3)− x+ 3

2ln |2x+ 3|+ C.

d) We replace u = ex, v′ = cosx. Then u′ = ex, v = sinx and∫ex cos dx = ex sinx−

∫ex sin dx.


This method we use again the second times. Now, we replace u = ex, v′ = sinx. Then u′ = ex, v = − cosx and∫ex cosx dx = ex sinx−

[ex(− cosx)−

∫ex(− cosx) dx

]=

= ex sinx+ ex cosx−∫ex cosx dx.

On the right side of the last equation we have the origin problem again. From this is possible to express finding indefinite integral:

2

∫ex cosx dx = ex sinx+ ex cosx,

∫ex cosx dx =

1

2ex(sinx+ cosx) + C. �

Problems

Using method by parts determine following indefinite integrals.67.

∫x lnx dx. 68.

∫x ex dx.

69.∫x e−x dx. 70.

∫x 3x dx.

71.∫x e2x dx. 72.

∫x ln2 x dx.

73.∫x arctg x dx. 74.

∫(9x2 + 4x) lnx dx.

75.∫x sinx dx. 76.

∫x cos 2x dx.

77.∫

xcos2 x

dx. 78.∫x tg2 x dx.

79.∫(x+ 1) ex dx. 80.

∫(x− 3) sinx dx.

81.∫x2 e−x dx. 82.

∫x2 sin 2x dx.

83.∫4x cos2 x dx. 84.

∫(x2 − 2) cosx dx.

85.∫(x2 + 2x) ex dx. 86.

∫(x2 + 6x+ 3) cos 2x dx.

87.∫x3 sinx dx. 88.

∫x3 arctg x dx.

89.∫lnx dx. 90.

∫arctg x dx.

91.∫arcsinx dx. 92.

∫(arcsinx)2 dx.

93.∫ln2 x dx. 94.

∫ln(x2 + 1) dx.

95.∫x ln(x2 + 3) dx. 96.

∫cos(lnx) dx.

97.∫earcsinx dx. 98.

∫ex arctg ex dx.

99.∫e2x sinx dx. 100.

∫ex sin2 x dx.

101.∫e3x sin 2x dx. 102.

∫e2x cos 5x dx.

103.∫

x√1−x2 arcsinx dx. 104.

∫x2

1+x2arctg x dx.


105.∫x cotg x

sin2 xdx. 106.

∫ln3 xx2

dx.

Indefinite integral – Integration of the partial fractions

Integration of the so called partial fractions is the preparatory phase of the solving of a much more complex problem, such as the integration of rationalfunctions.

As a partial fraction we call the rational function of the form

A

(x− α)nor

Mx+N

(x2 + px+ q)n,

where n is a natural number, A,α,M,N, p, q are any real number and the discriminant of the quadratic trinomial is negative (D < 0).A. Integration of the partial fraction A

(x−α)n .

a) If n = 1, then ∫A

x− αdx = A

∫1

x− αdx = A ln |x− α|+ C.

Excercise 5. Let’s solve∫

5x−7 dx.

Solution. ∫5

x− 7dx = 5

∫1

x− 7dx = 5 ln |x− 7|+ C. �

b) If n > 1, then∫

A(x−α)n dx we calculate by replacing x− α = z.

Excercise 6. Let’s solve∫

4(x+5)7

dx.

Solution. We introduce the substitution x+ 5 = z, dx = dz and we get∫4

(x+ 5)7dx =

∫4

z7dz = 4

∫z−7 dz = 4

z−6

−6+ C =

−23z6

+ C =−2

3(x+ 5)6+ C. �


B. Integration of the partial fractionMx+N

(x2 + px+ q)n. (3)

a) For n = 1 we have∫

Mx+Nx2+px+q

dx. This integral (3) is possible calculate by following way:

1) The trinomial x2 + px+ q modify to the form of full square, x2 + px+ q = (x+ p2)2 − (p

2)2 + q.

2) Next we introduce the substitution x+ p2= z

3) After simplifying the integral is possible to divide into two integrals, which each of them can be calculated usingknowing formulas or methods.

Excercise 7. Let’s calculate I =∫

4x−9x2−6x+13

dx.

Solution. D = 36− 52 = −16 < 0.

I =

∫4x− 9

x2 − 6x+ 13dx =

∫4x− 9

(x− 3)2 − 9 + 13dx.

After substitution x− 3 = z, x = z + 3, dx = dz we have

I =

∫4(z + 3)− 9

z2 + 4dz =

∫4z + 3

z2 + 4dz = 2

∫2z

z2 + 4dz + 3

∫1

z2 + 4dz.

I = 2 ln(z2 + 4)− 31√4arctg

z√4+ C = 2 ln(x2 − 6x+ 13)− 3

2arctg

x− 3

2+ C. �

b) If n > 1, is possible to apply a similar procedure as in case a) but in addition we need next recurrent formula

In =

∫1

(x2 + a2)ndx =

1

a2

[x

2(n− 1)(x2 + a2)n−1+

2n− 3

2n− 2In−1

]. (4)

Excercise 8. Let’s calculate I3 =∫

1(x2+4)3

dx.

Solution. Using last recurrent formula (4) we have

I3 =1

4

[x

2(3− 1)(x2 + 4)3−1+

6− 3

6− 2I2

]=

x

16(x2 + 4)2+

3

16I2.


I2 we calculate again using (4).

I2 =1

4

[x

2(2− 1)(x2 + 4)2−1+

4− 3

4− 2I1

]=

x

8(x2 + 4)+

1

8I1.

I1 =

∫1

x2 + 4dx =

1

2arctg

x

2.

After replacing to I1 and I2 we get

I3 =x

16(x2 + 4)2+

3

16

(x

8(x2 + 4)+

1

8

1

2arctg

x

2

)+ C =

=x

16(x2 + 4)2+

3x

128(x2 + 4)+

3

256arctg

x

2+ C. �


2x−5(x2+2x+2)2

dx.

Solution.

I =

∫2x− 5

[(x+ 1)2 − 1 + 2]2dx =

∫2x− 5

[(x+ 1)2 + 1]2dx.

We introduce the substitution x+ 1 = z, x = z − 1, dx = dz. Next:

I =

∫2(z − 1)− 5

(z2 + 1)2dz =

∫2z − 7

(z2 + 1)2dz =

∫2z dz

(z2 + 1)2dz − 7

∫dz

(z2 + 1)2.

The first integral we calculate by substitution z2 + 1 = t, 2zdz = dt.∫2z

(z2 + 1)2dz =

∫dt

t2= −1

t= − 1

z2 + 1= − 1

x2 + 2x+ 2.

The second integral we calculate using the recurrent formula (4) and we get∫1

(z2 + 1)2dz =

x+ 1

2(x2 + 2x+ 2)+

1

2arctg(x+ 1).

After re-substituting we get

I = − 1

x2 + 2x+ 2− 7

(x+ 1

2(x2 + 2x+ 2)+

1

2arctg(x+ 1)

)+ C =

= − 7x+ 9

x2 + x+ 1− 7

2arctg(x+ 1) + C. �


Problems

Determine following indefinite integrals of the partial fractions.107.

∫2

x+7dx. 108.

∫5

3−2x dx.

109.∫

12x−1 dx. 110.

∫ −83x+2

dx.

111.∫

4(4x−3)2 dx. 112.

∫1

4(2x+1)3dx.

113.∫

1x2−2x+5

dx. 114.∫

1x2+10x+34

dx.

115.∫

12x2+2x+5

dx. 116.∫

15x2+2x+10

dx.

117.∫

2xx2+6x+10

dx. 118.∫ −x

2x2−2x+1dx.

119.∫

12xx2+x+6

dx. 120.∫

3xx2+2x+10

dx.

121.∫

x+2x2+9

dx. 122.∫

2x−1x2+4

dx.

123.∫

10x+2x2−4x+5

dx. 124.∫

2−xx2−2x+5

dx.

125.∫

8(x2+1)2

dx. 126.∫

2x+6(x2+9)2

dx.

Indefinite integral – Integration of rational functions

Let us now solve fully the general problem of integrating of the rational functions. In this step is sufficient when we restrict the general problem to theproblem of integrating so called simple rational functions, where the degree of the numerator is less than the degree of the denominator. Now followsa fundamental point, which is the decomposition of simple rational functions of the sum of partial fractions. The whole procedure is described in thefollowing steps.

As a rational function f(x) is called the ratio of two polynomials Pm(x) and Pn(x)

f(x) =Pm(x)

Pn(x), (5)

where m and n are their degrees.If m < n, then the the rational function f(x) is called as a simple rational function. Now we specify the method as of a rational function, which

is not simple rational is possible get a simple rational and next how is possible the simple rational function decompose to the sum of partial fractions.Let’s consider a rational function f(x) that m ≥ n. In this case is necessary to divide the numerator by denominator and we get

Pm(x)

Pn(x)= Q(x) +

R(x)

Pn(x), (6)


where Q(x) is a polynomial and R(x) is a rest polynomial after dividing. Then polynomial R(x) has its degree less n, it means deg(R(x)) < n. ThusR(x)Pn(x)

is a simple rational function. In the next step by decomposition of the simple rational function R(x)Pn(x)

to the sum of partial fractions we proceed asfollows:

a) We find all real roots of the denominator Pn(x) and decompose it on the product of the numbers an (an of the coefficient at thehighest power of the polynomial Pn(x)) and next root factors of the real roots of a quadratic expressions whose discriminant isnegative (D < 0), (e.g. these quadratic expressions do not have real roots).

Pn(x) = an . . . (x− α)k . . . (x2 + px+ q)l . . .

Note that these different factors are the denominators of partial fractions.

b) An expression (x− α)k corresponds to k of the partial fractions in form:

A1

x− α,

A2

(x− α)2, . . . ,

Ak(x− α)k

.

An expression (x2 + px+ q)l corresponds to l of the partial fractions in form:

M1x+N1

x2 + px+ q,

M2x+N2

(x2 + px+ q)2, . . . ,

Mlx+Nl

(x2 + px+ q)l.

c) Then the rational function R(x)Pn(x)

is necessary to decompose to the sum of all partial fractions mentioned in previous step.

d) Obtained equality is multiplied by polynomial Pn(x) and we get the equality of two polynomials.

e) Then calculate coefficients Ai, Nj, Mj using one of the following methods:

a) method of comparing coefficients at the same powers,

b) substitute of (real) roots of the denominator,

c) combined method.

f) In the last step is possible to integrate the decomposed rational function at the one of the two form (5) or (6).


Excercise 10. Let’s calculate

a) I =

∫x2 − 2x+ 3

x2 + x− 2dx, b) I =

∫x2 − 8x− 2

x3 − 3x+ 2dx,

c) I =

∫x5 + x4 + 3x3 + x2 − 2

x4 − 1dx.

Solution.

a) (x2 − 2x+ 3) : (x2 + x− 2) = 1 + −3x+5x2+x−2

−(x2+x−2)−3x+5

I =

∫1 dx+

∫−3x+ 5

x2 + x− 2dx.

Function −3x+5x2+x−2 is a simple rational.

1. D = 9 > 0, so the denominator has two real roots. We get x2 + x− 2 = (x− 1)(x+ 2).

2. An expression x−1 corresponds to one partial fraction of the form Ax−1 . Similarly, an expression x+2 corresponds

to another partial fraction of the form Bx+2

.

3. we decompose the rational function to sum of two partial fractions:

−3x+ 5

x2 + x− 2=

A

x− 1+

B

x+ 2.

4. The equality is multiplied by x2 + x− 2 and we get −3x+ 5 = A(x+ 2) +B(x− 1).

5. Both coefficients, A and B we can calculate for example substitute of real roots of the denominator.

x = 1 : −3 + 5 = A(1 + 2) +B(1− 1), A = 23,

x = −2 : −3(−2) + 5 = A(−2 + 2) +B(−2− 1), B = −113.

6.

I = x+

∫ ( 23

x− 1+−11

3

x+ 2

)dx = x+

2

3ln |x− 1| − 11

3ln |x+ 2|+ C.

b) This is an integration of the simple rational function.


1. At first, we decompose the denominator. It is easy to find one root x1 = 1. Next we divide this polynomial byterm (x− 1). We have

(x3 − 3x+ 2) : (x− 1) = x2 + x− 2, x2 + x− 2 = (x− 1)(x+ 2).

Thenx3 − 3x+ 2 = (x− 1)(x− 1)(x+ 2) = (x− 1)2(x+ 2).

2. An expression (x−1)2 correspond two partial fractions of the forms A1

x−1 ,A2

(x−1)2 and an expression x+2 correspond

one fraction of the form Bx+2

.

3. Fully decomposition is possible write in this form:

x2 − 8x− 2

x3 − 3x+ 2=

A1

x− 1+

A2

(x− 1)2+

B

x+ 2.

4. Now, the last equality is multiplied by x3 − 3x+ 2 and we get

x2 − 8x− 2 = A1(x− 1)(x+ 2) + A2(x+ 2) +B(x− 1)2.

5. To calculate all coefficients A1, A2 and B we use the combined method. First we use the method of substitute

of real roots of the denominator. We havex = 1 : 12 − 8− 2 = A2(1 + 2), A2 = −3,x = −2 : (−2)2 − 8(−2)− 2 = B(−2,−1)2, B = 2.

To calculate all coefficients we use the method of comparing coefficients at the same powers. The right side wemodify and next compare coefficients at these powers.x2 : 1 = A1 +B,

1 = A1 + 2, A1 = −1.

I =

∫−1x− 1

dx+

∫−3

(x− 1)2dx+

∫2

x+ 2dx.

To the middle integral we introduce the substitution x− 1 = z, dx = dz and we get∫−3

(x− 1)2dx = −3

∫1

z2dz = −3

(−1

z

)=

3

z=

3

x− 1.


6. Finally,

I = − ln |x− 1|+ 3

x− 1+ 2 ln |x+ 2|+ C = ln

(x+ 2)2

|x− 1|+

3

x− 1+ C.

c) The degree of nominator is greater then denominator, so is necessary to divide.

(x5 + x4 + 3x3 + x2 − 2) : (x4 − 1) = x+ 1 +3x3 + x2 + x− 1

x4 − 1.

I =

∫x dx+

∫dx+

∫3x3 + x2 + x− 1

x4 − 1dx.

1. x4 − 1 = (x2 − 1)(x2 + 1) = (x− 1)(x+ 1)(x2 + 1).

2. Both expressions, x − 1 and x + 1 correspond partial fractions Ax−1 and B

x+1. Quadratic term x2 + 1 correspond

the partial fraction Mx+Nx2+1

.

3. Final form of decomposition is:

3x3 + x2 + x− 1

x4 − 1=

A

x− 1+

B

x+ 1+Mx+N

x2 + 1.

4. Last equality is multiplied by x4 − 1.

3x3 + x2 + x− 1 = A(x+ 1)(x2 + 1) +B(x− 1)(x2 + 1) + (Mx+N)(x2 − 1).

5. Coefficients A, B, M and N is possible to compute using combined method. First, we replace the roots ofdenominator.x = 1 : 3 · 1 + 1 + 1− 1 = A(1 + 1)(1 + 1), A = 1,x = −1 : 3 · (−1)3 + (−1)2 − 1− 1 = B(−1− 1)((−1)2 + 1), B = 1.

Compare the coefficients M a N in the powers, where are found, for exaple by x3 and x0.x3 : 3 = A+B +M, 3 = 1 + 1 +M, M = 1,x0 : −1 = A−B −N, −1 = 1− 1−N, N = −1.

6. Calculated coefficients substitute and complete calculation

I =x2

2+ x+

∫1

x− 1dx+

∫1

x+ 1dx+

∫x+ 1

x2 + 1dx.


∫x+ 1

x2 + 1dx =

∫x

x2 + 1dx+

∫1

x2 + 1dx =

1

2ln(x2 + 1) + arctg x.

I =x2

2+ x+ ln |x− 1|+ ln |x+ 1|+ 1

2ln(x2 + 1) + arctg x+ C =

=x2

2+ x+ ln

(|x2 − 1|

√x2 + 1

)+ arctg x+ C. �

Problems

Make a decomposition to sum of partial fractions.127. 2x

(x−1)(x−2) . 128. x−5x(x−2)(x+2)

.

129. 2x+11(x−1)2 . 130. 4−x2

x2(x−3) .

131. 7x−3x(x2+1)

. 132. 12(x−5)(x2+25)

.

133. −6x2(x2+4)

. 134. 1(x−1)(x2+2x+5)

.

135. x(x2+4)(x2−8x+17)

. 136. 3x(x2+1)2

.

In following problems calculate integrals of rational functions. The denominator contains only different real roots.137.

∫3x−4x2−2x dx. 138.

∫2

x2−1 dx.

139.∫

x+13x2+x−6 dx. 140.

∫3x−5

x2−3x+2dx.

141.∫

2x3+3x2+2x

dx. 142.∫

5x2x2−3x−2 dx.

143.∫

x2x2+3x+1

dx. 144.∫

84(x−1)(x−4)(x+3)

dx.

145.∫

6x2

x4−5x2+4dx. 146.

∫x+3x3−x dx.

147.∫

x2−5x+9x2−5x+6

dx. 148.∫

x2+2x−25x2+5x

dx.

149.∫

x3−x2−4x+12x2−4 dx. 150.

∫x3−7x−10x2−2x−3 dx.

151.∫

2x3

x2−1 dx. 152.∫ −3x4

x2+x−2 dx.

153.∫

5x3−15x2+15x−3x3−8x2+17x−10 dx. 154.

∫x3−14x3−x dx.

155.∫

x3−19x3−x dx. 156.

∫x5+x4−8x3−4x dx.

The denominator has only real roots, some are multiple.157.

∫2x+3

x2+2x+1dx. 158.

∫3x2+2x3−2x2 dx.

159.∫

x2+4x3+4x2+4x

dx. 160.∫

x2−3x+2x3+2x2+x

dx.

161.∫

x−1x3+5x2+8x+4

dx. 162.∫

x2

x3+5x2+8x+4dx.


163.∫

4x2+14x3+4x2+x

dx. 164.∫

x2−2x+3(x−1)(x3−4x2+3x)

dx.

165.∫

x2−8x+6x4−x3 dx. 166.

∫3x2+1(x2−1)2 dx.

167.∫

x3+4x4−4x3+4x2

dx. 168.∫

2x3+x+2x5+2x4

dx.

169.∫

x3+1x3−x2 dx. 170.

∫x3+x+2x3−2x2+x dx.

171.∫

x4+8x4−4x2 dx. 172.

∫x3−28

x3+3x2−4 dx.

173.∫

4x4−27x2+27x+94x3−12x2+9x

dx. 174.∫

3x3−x2+13x3−x2 dx.

175.∫

4x3−4x+4x3−x2−x+1

dx. 176.∫

x3−10x2+38x−50x3−10x2+25x

dx.The denominator has varios roots, some are complex numbers.177.

∫1

x3+xdx. 178.

∫ −x2x4−1 dx.

179.∫

1x3+1

dx. 180.∫

xx3−1 dx.

181.∫

8xx4−16 dx. 182.

∫18

x3+9xdx.

183.∫

4x−12x3+x2+3x−5 dx. 184.

∫20

x3−x2+4x−4 dx.

185.∫

x2−13x3−3x2+7x−5 dx. 186.

∫3x+6

x4+5x2+4dx.

187.∫

4x2

x4+10x2+9dx. 188.

∫x3−16x4+16x2

dx.

189.∫

x2+3x+2x2+2x+2

dx. 190.∫

x4

x2+3dx.

191.∫

x3−4x2+14x−13x3−4x2+13x

dx. 192.∫

x4−2x3+11x2+50x4−2x3+10x2

dx.

193.∫

x3−x2+10x3−x2−7x+15

dx. 194.∫

x4−3x3+20x2−x−18x4+5x2−36 dx.

195.∫

x4+1x3−x2+x−1 dx. 196.

∫x5+2x3+4x+4x4+2x3+2x2

dx.The denominator has various roots, some are multiple complex numbers.197.

∫12

x(x2+1)2dx. 198.

∫x4−x3+5x2−3x(x+2)(x2+1)2

dx.

199.∫

6(x−2)(x2−4x+5)2

dx. 200.∫

6x+1(x2+1)2

dx.

201.∫

x(x2+3x+3)2

dx. 202.∫

x+1x4+4x2+4

dx.

203.∫

4x(x+1)(1+x2)2

dx. 204.∫

x4+3x3

(x−1)(x2+1)2dx.

205.∫

1(x−1)(x2−2x+3)2

dx. 206.∫

1x4(x3+1)2

dx.

Indefinite integral – Integration of irrational functions

In this section we describe the procedures for integrating certain functions containing roots of linear, respectively. quadratic expressions.

A. Integral of the type∫R(x, n

√ax+ b) dx


This type of integral is possible modify using substitution ax+ b = zn to integral of rational function.


I =

∫2x+

√2x− 3

3 4√2x− 3 + 4

√(2x− 3)3

dx.

Solution. In this case n = 4. So we put substitution 2x− 3 = z4, x = z4+32

, dx = 124z3dz = 2z3dz.

I =

∫(z4 + 3 + z2)2z3

3z + z3dz = 2

∫z6 + z4 + 3z2

3 + z2dz.

Now we have known integral of rational function. It is necessary divide polynomials in this fraction. We get

z6 + z4 + 3z2

3 + z2= z4 − 2z2 + 9− 27

3 + z2.

I =z5

5− 2

z3

3+ 9z − 27 · 1√

3arctg

z√3+ C =

=4√

(2x− 3)5

5− 2

34√

(2x− 3)3 + 9 4√2x− 3− 27√

3arctg

4√2x− 3√

3+ C. �

B. Integral of the type∫R(x, n

√ax+bcx+d

) dx

Similarly, is possible modify this type using substitution ax+bcx+d

= zn again to integral of rational function.


I =

∫1

x

√x− 2

xdx.

Solution. In this case n = 2. So we put substitution x−2x

= z2, x− 2 = xz2, x(1− z2) = 2, x = 21−z2 , dx = 4z

(1−z2)2 dz. Then

I =

∫1− z2

2z

4z

(1− z2)2dz = 2

∫z2

1− z2dz = 2

∫z2 − 1 + 1

1− z2dz =

= 2

∫ (−1 + 1

1− z2

)dz = −2z + 2

∫1

1− z2dz.


Due to 1− z2 = (1− z)(1 + z), the function 11−z2 is necessary decompose to partial fractions.

1

1− z2=

A

1− z+

B

1 + z, 1 = A(1 + z) +B(1− z).

x = 1 : 1 = A(1 + 1), A = 12,

x = −1 : 1 = B(1 + 1), B = 12.∫1

1− z2dz =

∫ 12

1− zdz +

∫ 12

1 + zdz = −1

2ln |1− z|+ 1

2ln |1 + z| = 1

2ln |1 + z

1− z|.

I = −2z + ln |1 + z

1− z|+ C.

Finally, we come back to result origin variable, it means, replace z =√

x−2x. �

C. Integral of the type∫

Dx+E√Ax2+Bx+C

dxThis type of integral is possible calculate following way:

a) Ax2 +Bx+ C modify by full square method.

b) The term in brackets replace by z.

c) Function modify and divide the integral into two integrals, which have some form of type (α)− (δ).

Next calculations:

(α)

∫1√

a2 − x2dx, (β)

∫x√

a2 − x2dx, (γ)

∫1√

x2 + adx, (δ)

∫x√

x2 + adx.

The integral (α) calculate by substitution x = az. After editing we get∫1√

a2 − x2dx = arcsin

x

a.

Both integrals, (β) and (δ) calculate by substitution a2 − x2 = z2 and x2 + a = z2. Integral (γ) we calculate using basic rule (No. 11).


a) I =

∫x− 5√x2 + 10x

dx, b) I =

∫x+ 5√

6− 2x− x2dx.

Solution.


a) term x2 + 10x we modify to full square term and next use substitution

x2 + 10x = (x+ 5)2 − 25, x+ 5 = z, x = z − 5, dx = dz.

We get

I =

∫x− 5√

(x+ 5)2 − 25dx =

∫z − 5− 5√z2 − 25

dz =

∫z − 10√z2 − 25

dz =

=

∫z√

z2 − 25dz − 10

∫1√

z2 − 25dz.

The first integral is as (δ) and the second as (γ). Into first integral we put substitution z2 − 25 = t2, 2zdz = 2tdt and we have∫z√

z2 − 25dz =

∫tdt

t= t =

√z2 − 25.

ThenI =√z2 − 25− 10 ln |z +

√z2 − 25|+ C =

√x2 + 10x− 10 ln |x+ 5 +

√x2 + 10x|+ C. �

b) term 6− 2x− x2 we modify to full square term

I =

∫x+ 5√

7− (x+ 1)2dx =

∫z − 1 + 5√

7− z2dz =

∫z + 4√7− z2

dz =

=

∫z√

7− z2dz + 4

∫1√

7− z2dz.

The first integral is as (β) and the second as (α). Into first integral we put substitution 7 − z2 = t2, −2zdz = 2tdt, zdz = −tdtand we get ∫

z√7− z2

dz = −∫tdt

t= −t = −

√7− z2.

Finally

I = −√7− z2 + 4arcsin

z√7+ C = −

√6− 2x− x2 + 4arcsin

x+ 1√7

+ C. �


D. Method of undetermined coefficientsThis method is possible calculate integral of type

∫ Pn(x)√ax2+bx+c

dx, where Pn(x) is a polynomial of degree n ≥ 1. Then holds following equality:∫Pn(x)√

ax2 + bx+ cdx = Qn−1(x)

√ax2 + bx+ c+ k

∫1√

ax2 + bx+ cdx, (7)

where Qn−1(x) is a polynomial of degree n− 1 (with undetermined coefficients A,B,C, · · · and k as constant. These undetermined coefficients of thepolynomial Qn−1(x) and constant k is possible calculate by following process. The equality (7) derivative and obtained new equality multiply by term√ax2 + bx+ c. We get equality of two polynomials. Compare coefficients by same powers.

The integral on the right-hand side of equality (7) calculate using process described at point C.

Excercise 14. Let’s calculate I =∫x√x2 + 2x+ 2 dx.

Solution. This integral is not among the types listed above, but it can be adapted to such, if integrated function multiply and divide by√x2 + 2x+ 2.

We get

I =

∫x(x2 + 2x+ 2)√x2 + 2x+ 2

dx =

∫x3 + 2x2 + 2x√x2 + 2x+ 2

dx.

Using (7) we have ∫x3 + 2x2 + 2x√x2 + 2x+ 2

dx = (Ax2 +Bx+ C)√x2 + 2x+ 2 + k

∫1√

x2 + 2x+ 2dx.

After derivative we getx3 + 2x2 + 2x√x2 + 2x+ 2

= (2Ax+B)√x2 + 2x+ 2 + (Ax2 +Bx+ C)

2x+ 2

2√x2 + 2x+ 2

+

+k1√

x2 + 2x+ 2.

Multiplying by√x2 + 2x+ 2 we have

x3 + 2x2 + 2x = (2Ax+B)(x2 + 2x+ 2) + (Ax2 +Bx+ C)(x+ 1) + k.

Compare coefficients:x3 : 1 = 2A+ A, A = 1

3

x2 : 2 = B + 4A+B + A, B = 16

x1 : 2 = 2B + 4A+ C +B, C = 16

x0 : 0 = 2B + C + k, k = −12.


Thus

I = (1

3x2 +

1

6x+

1

6)√x2 + 2x+ 2− 1

2

∫1√

x2 + 2x+ 2dx.

The integral on the right-hand include the type A. The trinomial x2+2x+2 modify the full square, e.g. x2+2x+2 = (x+1)2− 1+2 = (x+1)2+1.After substitution x+ 1 = z, dx = dz we get∫

1√x2 + 2x+ 2

dx =

∫1√z2 + 1

dz = ln |z +√z2 + 1| = ln |x+ 1 +

√x2 + 2x+ 2|.

Finally,

I =1

3

(x2 +

1

2x+

1

2

)√x2 + 2x+ 2− 1

2ln |x+ 1 +

√x2 + 2x+ 2|+ C. �

Problems

A.∫R(x, n

√ax+ b) dx

207.∫

1−√x

1+√xdx. 208.

∫ √x−4

x+4−4√xdx.

209.∫ √

x3√x+1

dx. 210.∫ 3√x

x+6√x5dx.

211.∫ √

x+ 4√x+ 3√x2(x+

6√x7)

dx. 212.∫ 6√x+1

6√x7+

4√x5dx.

213.∫

1x√x−4 dx. 214.

∫x+1

3√3x+1dx.

215.∫

x√x+1+ 3√x+1

dx. 216.∫ 3√x−1

2√x−1− 3√

(x−1)2dx.

B.∫R(x, n

√ax+bcx+d

)dx

217.∫ √

1−x1+x· 1xdx. 218.

∫ √1+x1−x

1(1−x)(1+x)2 dx.

219.∫ √

1−x1+x

dx. 220.∫

1x

√xx−1 dx.

C.∫

Dx+E√Ax2+Bx+C

dx

221.∫

1√1−2x−x2 dx. 222.

∫3x+1√

x2+5x−10 dx.

223.∫

x+3√x2+2x

dx. 224.∫

1√12x−9x2−2 dx.

225.∫

1√x2+x+1

dx. 226.∫

1√9x2−6x+2

dx.

227.∫

1√5x2+8x−3 dx. 228.

∫1√

3x2−7x+5dx.

229.∫

2x−10√1+x−x2 dx. 230.

∫8x−11√5+2x−x2 dx.


D.∫ Pn(x)√

ax2+bx+cdx

231.∫

2x2−3x√x2−2x+5

dx. 232.∫

x2√x2+2x+2

dx.

233.∫

x3√1−2x−x2 dx. 234.

∫x3+5x2+8x+3√

x2+4x+3dx.

235.∫

x2√1−2x−x2 dx. 236.

∫ √x2 + 4x+ 3 dx.

237.∫ √

x2 + 4x+ 13 dx. 238.∫ √

1− 2x− 3x2 dx.

239.∫ √

3− 2x− x2 dx. 240.∫(4x− 10)

√2 + 3x− x2 dx.

Indefinite integral – Integration of trigonometric functions

In the process of integration of trigonometric functions are often used substitution method, which usually prevent some modifications of the integralfunctions using several well-known formulas valid for trigonometric functions and relations between them. In solving certain specific types of problemscan also be used some recurrent formulas.

A.∫sinn x dx,

∫cosn x dx, n is integer

These types of integrals is possible to calculate using following recurrent formulas

In =

∫sinn x dx = − 1

ncosx sinn−1 x+

n− 1

nIn−2, (8)

In =

∫cosn x dx =

1

nsinx cosn−1 x+

n− 1

nIn−2. (9)

Re-using these formulas is possible to reach the integral I0 or I1 (in case (8) I0 =∫sin0 x dx =

∫dx = x, I1 =

∫sinx dx = − cosx; in case (9)

analogically).


a)

∫sin5 x dx, b)

∫cos4 x dx.

Solution.

a) We use the recurrent formula (8) for n = 5 and next again for n = 3

I5 =

∫sin5 x dx = −1

5cosx sin4 x+

4

5I3 = −

1

5cosx sin4 x+

4

5

(−1

3cosx sin2 x+

2

3I1

)=


= −1

5cosx sin4 x− 4

15cosx sin2 x+

8

15

∫sinx dx =

= −1

5cosx sin4 x− 4

15cosx sin2 x− 8

15cosx+ C.

b) We use the recurrent formula (9) for n = 4 and next again for n = 2

I4 =1

4sinx cos3 x+

3

4I2 =

1

4sinx cos3 x+

3

4

(1

2sinx cosx+

1

2I0

)=

=1

4sinx cos3 x+

3

8sinx cosx+

3

8

∫cos0 x dx =

1

4sinx cos3 x+

3

8sinx cosx+

3

8x+ C. �

Note that integrals of this type can be also calculate without the use of recurrent formulas.

a) if n is odd, the function modify and introduce appropriate substitution.

b) If n is even, we reduce the power of successively applying double angle formulas for both, sine and cosine functions

sin2 α =1− cos 2α

2, cos2 α =

1 + cos 2α

2.

Let’s calculate again Excercise 15 using last methods.

a) ∫sin5 x dx =

∫sin4 x sinx dx =

∫(sin2 x)2 sinx dx =

∫(1− cos2 x)2 sinx dx.

We use substitution cosx = z, − sinx dx = dz, sinx dx = −dz. Then∫sin5 x dx = −

∫(1− z2)2dz = −

∫(1− 2z2 + z4)dz = −(z − 2

z3

3+z5

5) + C =

= − cosx+2

3cos3 x− 1

5cos5 x+ C.


b) ∫cos4 x dx =

∫(cos2 x)2 dx =

∫ (1 + cos 2x

2

)2

dx =1

4

∫(1 + 2 cos 2x+ cos2 2x) dx =

=1

4(x+ sin 2x+

∫cos2 2x dx) =

1

4x+

1

4sin 2x+

1

4

∫1 + cos 4x

2dx =

=1

4x+

1

4sin 2x+

1

8

(x+

1

4sin 4x

)+ C =

3

8x+

1

4sin 2x+

1

32sin 4x+ C. �

B. Integrals of types∫sinn x cosm x dx, n, m are integer

There are two cases:

a) both integer numbers, n and m are even. Then is possible modify the function this way, that we get only powers of functions sinxor cosx. These integrals are calculated using recurrent formulas.

Excercise 16. Let’s calculate I =∫sin4 x cos6 x dx.

Solution. It is preferable to modify the power of the smaller exponent, it means sin4 x.

I =

∫(sin2 x)2 cos6 x dx =

∫(1− cos2 x)2 cos6 x dx =

∫(1− 2 cos2 x+ cos4 x) cos6 x dx =

=

∫cos6 x dx− 2

∫cos8 x dx+

∫cos10 x dx.

Each of these integrals can be calculated using the recursive formula (9). �

b) At least one of the numbers n,m is odd. If n (m) is odd, after editing is possible use substitution cosx = z (sinx = z).

Excercise 17. Let’s calculate I =∫sin2 x cos5 x dx.

Solution. Now, it is possible to use similarly modification as in Problem 15a) (with the difference that now modify cos5 x).

I =

∫sin2 x cos4 x cosx dx =

∫sin2 x(1− sin2 x)2 cosx dx.


After substituting sinx = z, cosx dx = dz we have

I =

∫z2(1− z2)2dz =

∫(z2 − 2z4 + z6)dz =

z3

3− 2

z5

5+z7

7+ C =

=1

3sin3 x− 2

5sin5 x+

1

7sin7 x+ C. �

C. Integrals of types∫R(sinx) cosx dx,

∫R(cosx) sinx dx

These indefinite integrals is preferable to calculate using substitution (after first rule about substitution)

sinx = z, resp. cosx = z.


a) I =

∫1

sin2 x cos3 xdx, b) I =

∫sin3 x

cos4 xdx.

Solution.

a) First, the origin function modify by following way:

I =

∫1

sin2 x cos4 xcosx dx =

∫1

sin2 x(1− sin2 x)2cosx dx.

Last integral include to first type and we use substitution sinx = z, cosx dx = dz. We get

I =

∫1

z2(1− z2)2dx.

We got known integral of rational function, which we know decompose to partial fractions and finalize calculation (make in-house).

b)

I =

∫sin2 x

cos4 xsinx dx =

∫1− cos2 x

cos4 xsinx dx.

Last integral include to second type and we use substitution cosx = z, − sinx dx = dz and we have

I = −∫

1− z2

z4dz = −

∫ (1

z4− 1

z2

)dz =

1

3z3− 1

z+ C =

1

3 cos3 x− 1

cosx+ C. �


D. Indefinite integrals of type∫R(sinx, cosx) dx

Indefinite integrals of this type is possible modify using substitution tg x2= z to indefinite integral of rational function. Functions sinx, cosx a dx we

modify using new variable z. Plat

sinx =2z

1 + z2, cosx =

1− z2

1 + z2, dx =

2

1 + z2dz.


dx4 cosx+3 sinx

.

Solution. We use substitution tg x2= z and we have

I =

∫ 21+z2

dz

41−z21+z2

+ 3 2z1+z2

= −∫

1

2z2 − 3z − 2dz = −

∫1

(z − 2)(2z + 1)dz,

1

(z − 2)(2z + 1)=

A

z − 2+

B

2z + 1.

Using elementary method of finding roots we get z = 2 and z = −12. Then A = 1

5, B = −2

5and we have

I = −(1

5ln |z − 2| − 2

5

1

2ln |2z + 1|

)+ C =

1

5ln |

2 tg x2+ 1

tg x2− 2|+ C. �

If in some integral we can found only even powers of functions sinx and cosx, then we use the substitution tg x = z. Thus

sin2 x =z2

1 + z2, cos2 x =

1

1 + z2, dx =

1

1 + z2dz.


3+sin2 x2 cos2 x−cos4 x dx.

Solution. We use substitution tg x = z and we have

I =

∫3 + z2

1+z2

2 11+z2− 1

(1+z2)2

1

1 + z2dz =

∫4z2 + 3

2z2 + 1dz.

After dividing the numerator of the denominator we have

I =

∫ (2 +

1

2z2 + 1

)dz = 2z +

1

2

∫1

z2 + 12

dz = 2z +

√2

2arctg(

√2z) + C =


= 2 tg x+

√2

2arctg(

√2 tg x) + C. �

Integrals A-C are also of type∫R(sinx, cosx) dx. Although we do not solve them by substituting tg x

2= z, because its use leads to complicated

calculations.E. Use trigonometric substitution to calculate integrals of irrational functionsIndefinite integrals of types: ∫

R(x,√a2 − x2) dx,

∫R(x,

√a2 + x2) dx,

∫R(x,

√x2 − a2) dx

is possible also modify using so called goniometric substitutions

x = a sin z, x = a tg z, x =a

sin z

to known type∫R(sinx, cosx) dx.


a) I =

∫dx

(√x2 + 4)3

, b) I =

∫ √−x2 + 10x− 16 dx.

Solution.

a) This integral include to first type and, a = 2. Therefore, we use the substitution x = 2 tg z, dx = 2cos2 z

dz,

x2 + 4 = 4 tg2 z + 4 = 4(tg2 z + 1) = 4

(sin2 z

cos2 z+ 1

)= 4

sin2 z + cos2 z

cos2 z=

4

cos2 z.

Then

I =

∫ 2cos2 z

dz(2

cos z

)3 =1

4

∫cos zdz =

1

4sin z + C.

We come back to origin variable x. We don’t express z, but sin z.

tg z =x

2,

sin z =tg z√

1 + tg2 z=

x2√

1 + x2

4

=x√x2 + 4

.


Then

I =1

4

x√x2 + 4

+ C. �

b) The expression under the square root we will add to the full square.

−x2 + 10x− 16 = −(x2 − 10x+ 16) = −[(x− 5)2 − 25 + 16] = 9− (x− 5)2.

Then

I =

∫ √9− (x− 5)2 dx.

Now we use the substitution x− 5 = z, dx = dz and we have

I =

∫ √9− z2dz.

Last integral include to first type and a = 3. Therefore, we use the substitution z = 3 sin t, dz = 3 cos tdt, 9 − z2 = 9 − 9 sin2 t =9(1− sin2 t) = 9 cos2 t. Thus

I = 9

∫cos2 tdt = 9

∫1 + cos 2t

2dt =

9

2

(t+

1

2sin 2t

)+ C.

Because sin t = z3

then

cos t =√1− sin2 t =

√1− z2

9=

√9− z23

, t = arcsinz

3,

sin 2t = 2 sin t cos t = 2z

3

√9− z23

=2

9z√9− z2.

Finally we get

I =9

2

(arcsin

z

3+

1

9z√9− z2

)+ C =

=9

2

(arcsin

x− 5

3+

1

9(x− 5)

√−x2 + 10x− 16

)+ C. �


Problems

A.∫sinn x dx,

∫cosn x dx

241.∫sin2 x dx. 242.

∫cos2 x dx.

243.∫sin3 x dx. 244.

∫cos3 x dx.

245.∫sin4 x dx. 246.

∫cos4 x dx.

247.∫sin7 x dx. 248.

∫cos5 x dx.

B.∫sinn x cosm x dx.

249.∫sinx cos2 x dx. 250.

∫sin2 x cos3 x dx.

251.∫sin3 x cos2 x dx. 252.

∫sin7 x cosx dx.

253.∫sin5 x cos2 x dx. 254.


255.∫sin3 x cos5 x dx. 256.


257.∫sin4 x cos2 x dx. 258.

∫sin4 x · cos4 x dx.

C.∫R(sinx) cosx dx,

∫R(cosx) sinx dx

259.∫

sinxcos2 x

dx. 260.∫

sin3 xcos2 x

dx.

261.∫

sin3 xcos3 x

dx. 262.∫

sin3 xcos4 x

dx.

263.∫

cosxsin5 x

dx. 264.∫

cos5 xsin2 x

dx.

265.∫

sin2 xcos3 x

dx. 266.∫

cos2 xsinx

dx.

267.∫

cos4 xsinx

dx. 268.∫

sin2 xcosx

dx.

269.∫

sin2 xcos2 x

dx. 270.∫

sin4 xcos2 x

dx.271.

∫cosx

sin2 x+9dx. 272.

∫cosx

sin2 x+6 sinx+5dx.

273.∫

sinxcos2 x+2 cosx

dx. 274.∫

sinxcos2 x−2 cosx+1

dx.

275.∫

sin3 x2+cosx

dx. 276.∫

cosx(1−sinx)3 dx.

277.∫

1cos3 x

dx. 278.∫

1cosx

dx.279.

∫1

sin2 x cosxdx. 280.

∫1

sinx cosxdx.

D.∫R(sinx, cosx) dx

281.∫

15−3 cosx dx. 282.

∫1

5−4 sinx+3 cosxdx.

283.∫

2−sinx2+cosx

dx. 284.∫

x+sinx1+cosx

dx.

285.∫

19+4 cosx

dx. 286.∫

cosx1+cosx

dx.

287.∫

sinxsinx+cosx

dx. 288.∫

1+sinx1−sinx dx.

289.∫

18−4 sinx+7 cosx

dx. 290.∫

1cosx+2 sinx+3

dx.

291.∫

1−tg x1+tg x

dx. 292.∫

11+8 cos2 x

dx.


293.∫

14 sin2 x+9 cos2 x

dx. 294.∫

14−3 cos2 x+5 sin2 x

dx.

295.∫

11−3 sinx·cosx−5 cos2 x dx. 296.

∫cosx

sin3 x−cos3 x dx.

297.∫

1sin2 x−5 sinx·cosx dx. 298.

∫1

sin2 x−tg2 x dx.

299.∫

1cos4 x

dx. 300.∫

1sin4 x

dx.

E.∫R(x,

√a2 − x2) dx,

∫R(x,

√a2 + x2) dx,

∫R(x,

√x2 − a2) dx

301.∫

1x2√2+x2

dx. 302.∫

1x√a2+x2

dx.

303.∫ √

x2−9x

dx. 304.∫

x2√x2−4 dx.

305.∫ √

4−x2x2

dx. 306.∫

1√(1−x2)3

dx.

307.∫

1√(9+x2)3

dx. 308.∫

1x2√x2−9 dx.

Indefinite integral – Next different problems

Earlier in this chapter, we had the opportunity to meet with various ways and methods of calculation of indefinite integrals. That’s when you use such amethod, we learn the only way to calculate the number of different tasks. We must learn to ably use a variety of methods and appropriate combinationsof them. Also important is the modification of functions to integrate, if so required by the procedure. It often happens that in one example we mustuse more methods.

Calculate following indefinite integrals.309.

∫ex−1ex+1

dx. 310.∫

ex−2e2x+4

dx.

311.∫

2e3x+3ex

e2x+1dx. 312.

∫e4x

e8x+4dx.

313.∫

1ax+1

dx. 314.∫

e3x+ex

e4x−e2x+1dx.

315.∫(ln3 x+ lnx) dx. 316.

∫arcsinxx2

dx.

317.∫

lnxx(1−ln2 x) dx. 318.

∫x·arctg x√

1+x2dx.

319.∫

1√1−x2 arccos2 x dx. 320.

∫ √1−√x dx.

321.∫ √

1−ex1+ex

dx. 322.∫

sinx3√1+2 cosx

dx.

323.∫

arctg xx4

dx. 324.∫

1sin 2x−2 sinx dx.

325.∫

1sin4 x+cos4 x

dx. 326.∫

1√1+ex+e2x

dx.

327.∫

1earcsin x

√1−x2 dx. 328.

∫ex√arctg ex

1+e2xdx.

329.∫

ln arctg x(1+x2) arctg x

dx. 330.∫ √

arccosx1−x2 dx.


Self-assessment questions

1. What is the difference between the concepts of antiderivative and indefinite integral?

2. Why is the process of integration in general, much harder than the process of derivation, which is actually the opposite procedure?

3. List three basic uses methods by parts.

4. Try to find at least three elementary functions whose indefinite integral can not be expressed by elementary functions.

5. Determine the types of elementary functions, for which we use to integrating process some recurrent formula.

Bibliography

[1] Siagova, J.: Advanced Mathematics, Bratislava, STU, 2011.

[2] Bubenik, F.: Problems to Mathematics for Engineers, Praha, CVUT, 1999.


[4] Neustupa, J.: Mathematics I, Praha, CVUT, 1996.

[5] Neustupa, J.: Mathematics II, Praha, CVUT, 1998.

Solutions

Answers to self-assessment questions (if you are not able to formulate this answers), as well as many other answers can be found in the recommendedliterature.

1. x3 + x2 + x+ 13 ln |x|+ C 2. 11

6 x6 − 4x3 + C

3. 2√x+ 2x+ 2

3x√x+ C 4. 2

√x− 4 4

√x+ C

5. ln |x| − 1x2 + C 6. 1

2x4 − 5x2 + 6 ln |x|+ C

7. ex + 1x + C 8. ax

ln a + 13x3 + C

9. 10x

ln 10 + C 10. 4x

ln 4 + 2 6x

ln 6 + 9x

ln 9 + C

11. 3( 13 )

x

ln 13

− 2( 43 )

x

ln 43

+ C 12.( 34 )

x

ln 34

− 2x+( 43 )

x

ln 43

+ C

13.( 45 )

x

ln 45

− 2x+( 54 )

x

ln 54

+ C 14. 5 sinx− 2√3

7

√x7 + 3arctg x+ C

15. tg x− x+ C 16. − cotg x− x+ C


17. − cotg x− tg x+ C 18. tg x− cotg x+ C19. 1

2x−12 sinx+ C 20. 1

2x+ 12 sinx+ C

21. sinx− cosx+ C 22. 12 tg x+ 1

2x+ C23. x− arctg x+ C 24. x− 2 arctg x+ C25. −x+ 5arctg x+ C 26. 1

3x3 − x+ arctg x+ C

27. 14x

4 − 2x2 + arctg x2 + C 28. x− 1

x − 2 ln |x|+ C29. 1

2 ln |x2 − 3|+ C 30. 1

3 ln |x3 + 1|+ C

31. 13 ln |3x

2 − 15x+ 22|+ C 32. ln | sinx| − ln | cosx|+ C33. − 1

2 ln |2 cos2 x+ 3|+ C 34. − 1

6 ln |1− 3e2x|+ C35. ln | lnx|+ C 36. ln | arcsinx|+ C

37. 110 (x+ 2)

10+ C 38. 2

3

√(x+ 3)

3+ C

39. − ln |5− x|+ C 40. 56 ln |6x− 1|+ C

41. − 2x+3 + C 42. − 1

3(2x+3)3 + C

43. 124 (x

2 − 4)12 + C 44. − 12 cos 2x+ C

45. 3 sin x3 + C 46. − 1

4 ln | cos(4x− 1)|+ C47. 1

3 arctg 3x+ C 48. 12 arcsin 2x+ C

49. −e−x + C 50. 12e

2x + C

51. 2√ex + C 52. 1

3ex3

+ C

53. 16 arctg

x3

2 + C 54. 12(1−x)2 −

21−x − ln |1− x|+ C

55. 23 tg(x

3 + 1) + C 56. 13 sin

3 x+ C

57. − 43

4√cos3 x+ C 58. ln | lnx|+ 2

ln x + C

59. arcsinx−√1− x2 + C 60. tg x− 2

cos x + C

61. 12 arcsin

2 x+√1− x2 + C 62. 1

2 ln2 cos x+ C

63. −e 1x + C 64. 2e

√x + C

65. 1ln 2 arcsin 2

x + C 66. 2√tg x− 1 + C

67. 12x

2 ln |x| − 14x

2 + C 68. xex − ex + C

69. −xe−x − e−x + C 70. 1ln 3x.3

x − 1ln2 3

3x + C

71. 12xe

2x − 14e

2x + C 72. 12x

2 ln2 x− 12x

2 ln |x|+ 14x

2 + C

73. 12

(x2 + 1

)arctg x− 1

2x+ C 74. (3x3 + 2x2) lnx− x3 − x2 + C75. −x cosx+ sinx+ C 76. 1

2x sin 2x+ 14 cos 2x+ C

77. x tg x+ ln | cosx|+ C 78. x tg x+ ln | cosx| − 12x

2 + C

79. xex + C 80. (3− x) cosx+ sinx+ C81. −e−x

(x2 + 2x+ 2

)+ C 82. − 1

4

(2x2 − 1

)cos 2x+ 1

2x sin 2x+ C83. 2x2 cos2 x− (x2 − 1

2 ) cos 2x+ x sin 2x+ C 84. (x2 − 2) sinx+ 2x cosx− 2 sinx+ C

85. x2ex + C 86. 12

(x2 + 6x+ 5

2

)sin 2x+ 1

2 (x+ 3) cos 2x+ C87. (−x3 + 6x) cosx+ (3x2 − 6) sinx+ C 88. 1

4 (x4 − 1) arctg x− 1

12x3 + 1

4x+ C

89. x lnx− x+ C 90. x arctg x− 12 ln(1 + x2) + C

91. x arcsinx+√1− x2 + C 92. x arcsin2 x+ 2

√1− x2 arcsinx− 2x+ C


93. x ln2 x− 2x lnx+ 2x+ C 94. x ln(x2 + 1)− 2x+ 2arctg x+ C95. 1

2 (x2 + 3) ln

(x2 + 3

)− 1

2 (x2 + 3) + C 96. 1

2x cos(ln |x|) +12x sin(ln |x|) + C

97. 12

(x+√1− x2

)earcsin x + C 98. ex arctg ex − 1

2 ln(1 + e2x) + C

99. 15e

2x(2 sinx− cosx) + C 100. ex(sin2 x− 1

5 sin 2x+ 25 cos 2x

)+ C

101. 113e

3x(3 sin 2x− 2 cos 2x) + C 102. 129e

2x (5 sin 5x+ 2 cos 5x) + C

103. x−√1− x2 arcsinx+ C 104. x arctg x− 1

2 arctg2 x− 1

2 ln(1 + x2) + C

105. − 12x cotg

2 x− 12 cotg x−

12x+ C 106. − 1

x

(ln3 x+ 3 ln2 x+ 6 lnx+ 6

)+ C

107. 2 ln |x+ 7|+ C 108. − 52 ln |3− 2x|+ C

109. 12 ln |2x− 1|+ C 110. − 8

3 ln |3x+ 2|+ C

111. − 14x−3 + C 112. − 1

16(2x+1)2 + C

113. 12 arctg

x−12 + C 114. 1

3 arctgx+53 + C

115. 13 arctg

2x+13 + C 116. 1

7 arctg5x+1

7 + C117. ln(x2 + 6x+ 10)− 6 arctg(x+ 3) + C 118. − 1

4 ln(x2 − x+ 1

2 )−12 arctg(2x− 1) + C

119. 6 ln(x2 + x+ 6)− 12√23

arctg 2x+1√23

+ C 120. 32 ln(x

2 + 2x+ 10)− arctg x+13 + C

121. 12 ln(x

2 + 9) + 23 arctg

x3 + C 122. ln(x2 + 4)− 1

2 arctgx2 + C

123. 5 ln(x2 − 4x+ 5) + 22 arctg(x− 2) + C 124. − 12 ln(x

2 − 2x+ 5) + 12 arctg

x−12 + C

125. 4xx2+1 + 4arctg x+ C 126. x−3

3(x2+9) +19 arctg

x3 + C

127. Ax−1 + B

x−2 128. Ax + B

x−2 + Cx+2

129. Ax−1 + B

(x−1)2 130. Ax + B

x2 + Cx−3

131. Ax + Bx+C

x2+1 132. Ax−5 + Bx+C

x2+25

133. Ax + B

x2 + Cx+Dx2+4 134. A

x−1 + Bx+Cx2+2x+5

135. Ax+Bx2+4 + Cx+D

x2−8x+17136. A

x + Bx+Cx2+1 + Dx+E

(x2+1)2

137. 2 ln |x|+ ln |x− 2|+ C 138. ln |x− 1| − ln |x+ 1|+ C139. 3 ln |x− 2| − 2 ln |x+ 3|+ C 140. 2 ln |x− 1|+ ln |x− 2|+ C141. ln |x| − 2 ln |x+ 1|+ ln |x+ 2|+ C 142. 2 ln |x− 2|+ 1

2 ln |2x+ 1|+ C143. ln |x+ 1| − 1

2 ln |2x+ 1|+ C 144. −7 ln |x− 1|+ 4 ln |x− 4|+ 3 ln |x+ 3|+ C145. 2 ln |x− 2| − ln |x− 1|+ ln |x+ 1| − 2 ln |x+ 2|+ C 146. −3 ln |x|+ 2 ln |x− 1|+ ln |x+ 1|+ C147. x+ 3 ln |x− 3| − 3 ln |x− 2|+ C 148. x− 5 ln |x|+ 2 ln |x+ 5|+ C149. 1

2x2 − x+ 2 ln |x− 2| − 2 ln |x+ 2|+ C 150. 1

2x2 + 2x− ln |x− 3|+ ln |x+ 1|+ C

151. x2 + ln |x− 1|+ ln |x+ 1|+ C 152. −x3 + 32x

2 − 9x+ ln (x+2)16

|x−1| + C

153. 5x+ ln∣∣ (x−1) 1

2 (x−5)1616

(x−2)73

∣∣+ C 154. x4 + ln |x| − 1

16 ln |(2x− 1)7(2x+ 1)9|+ C

155. x9 + ln |x| − 1

27 ln |(3x− 1)13(3x+ 1)14|+ C 156. x3

3 + x2

2 + 4x+ ln∣∣x2(x−2)5

(x+2)3

∣∣+ C

157. 2 ln |x+ 1| − 1x+1 + C 158. − 1

2 ln |x|+72 ln |x− 2|+ 1

x + C

159. 2 ln |x|+ ln |x+ 2|+ 4x+2 + C 160. 2 ln |x| − ln |x+ 1|+ 6

x+1 + C

161. 2 ln |x+ 2| − 2 ln |x+ 1| − 3x+2 + C 162. ln |x+ 1|+ 4

x+2 + C


163. ln |x|+ 22x+1 + C 164. ln

√(x−1)(x−3)|x| + 1

x−1 + C

165. ln |x| − ln |x− 1|+ 3x2 − 2

x + C 166. 12 ln |x− 1| − 1

2 ln |x+ 1| − 1x−1 −

1x+1 + C

167. ln |x| − 3x−2 −

1x + C 168. ln |x| − ln |x+ 2| − 1

3x3 + C

169. 2 ln |x− 1| − ln |x|+ x+ 1x + C 170. 2 ln |x|+ x− 4

x−1 + C

171. 32 ln |

x−2x+2 |+ x+ 2

x + C 172. −3 ln |x− 1|+ x− 12x+2 + C

173. ln | x2x−3 |+

12x

2 + 3x− 32x−3 + C 174. 3 ln |3x− 1| − 3 ln |x|+ x+ 1

x + C

175. 3 ln |x− 1|+ ln |x+ 1|+ 4x− 2x−1 + C 176. 2 ln |x− 5| − 2 ln |x|+ x− 3

x−5 + C

177. ln |x|√x2+1

+ C 178. 14 ln

∣∣x+1x−1

∣∣− 12 arctg x+ C

179. 16 ln

(x+1)2

x2−x+1 + 1√3arctg 2x−1√

3+ C 180. 1

3 ln|x−1|√x2+x+1

+ 1√3arctg 2x+1√

3+ C

181. 12 ln

x2−4x2+4 + C 182. ln x2

x2+9 + C

183. ln√x2+2x+5|x−1| + 3arctg x+1

2 + C 184. 4 ln |x− 1| − 2 ln(x2 + 4)− 2 arctg x2 + C

185. ln (x2−2x+5)2

|(x−1)3| + arctg x−12 + C 186. 1

2 lnx2+1x2+4 + 2arctg x− arctg x

2 + C

187. − 12 arctg x+ 3

2 arctgx3 + C 188. 1

x + 12 ln(x

2 + 16) + 14 arctg

x4 + C

189. x+ 12 ln(x

2 + 2x+ 2)− arctg(x+ 1) + C 190. x3

3 − 3x+ 3√3 arctg x√

3+ C

191. x+ 12 ln

x2−4x+13x2 − 1

3 arctgx−23 + C 192. x− 5

x + 12 ln

x2

x2−2x+10 − arctg x−13 + C

193. x− ln |x+ 3|+ 12 ln(x

2 − 4x+ 5) + 2 arctg(x− 2) + C 194. x+ ln |x− 2| − 2 ln |x+ 2| − ln(x2 + 9) + 3 arctg x3 + C

195. 12x

2 + x+ ln |x−1|√x2+1

− arctg x+ C 196. x2

2 − 2x− 2x + 2 ln(x2 + 2x+ 2)− 2 arctg(x+ 1) + C

197. 6x2+1 − 6 ln(x2 + 1) + 12 lnx+ C 198. x

x2+1 + 12 ln(x

2 + 1)− 2 ln(x+ 2) + C

199. 3x2−4x+5 + 3 ln (x−2)2

(x2−4x+5) + C 200. x−62(x2+1) +

12 arctg x+ C

201. − x+2x2+3x+3 −

2√3arctg 2x+3√

3+ C 202. − 1

2(x2+2) +x

4(x2+2) +1

4√2arctg x√

2+ C

203. ln√x2+1|x+1| + x−1

x2+1 + C 204. ln |x− 1|+ 2arctg x+ 2x−12(x2+1) + C

205. 14 ln

|x−1|√x2−2x+3

− 14(x2−2x+3) + C 206. 2

3 ln∣∣x3+1

x3

∣∣− 13x3 − 1

3(x3+1) + C

207. −x+ 4√x− 4 ln |

√x+ 1|+ C 208. 2

√x+ 8√

x−2 + C

209. 66√x7

7 − 66√x5

5 + 2√x− 6 6

√x+ 6arctg 6

√x+ C 210. 3

(6√x2 − 2 6

√x+ 2 ln

∣∣ 6√x− 1

∣∣)+ C

211. 32

3√x+ 6 12

√x− 6 arctg 12

√x+ C 212. 2 ln

12√x

1+ 12√x+ 1

12√x− 1

2 6√x+ C

213. arctg√x−42 + C 214. 1

153√(3x+ 1)5 + 1

33√(3x+ 1)2 + C

215. 6

(√(x+1)3

9 −3√

(x+1)4

8 +6√

(x+1)7

7 − x+16 +

6√

(x+1)5

5 −6√

(x+1)4

4

)+ C 216. − 18

5 t5 − 9t4 − 24t3 − 72t2 − 288t− 576 ln |t− 2|+ C, kde t = 6√x− 1

217. ln∣∣√1−x−

√1+x√

1−x+√1+x

∣∣+ 2arctg√

1−x1+x + C 218. x√

1−x2+ C

219.√1− x2 − arctg

√1−x1+x + C 220. ln(2x− 1 + 2

√x(x+ 1)) + C

221. arcsin x+1√2+ C 222. 3

√x2 + 5x− 10− 13

2 ln∣∣x+ 5

2 +√x2 + 5x− 10

∣∣+ C


223.√x2 + 2x+ 2 ln

∣∣x+ 1 +√x2 + 2x

∣∣+ C 224. 13 arcsin

3x−2√2

+ C

225. ln∣∣x+ 1

2 +√x2 + x+ 1

∣∣+ C 226. 13 ln |3x− 1 +

√9x2 − 6x+ 2|+ C = 1

3 ln |x−x−13 +

√x2 − 2

3x+ 29 |+ C

227. 1√5

∣∣x+ 45 +

√x2 + 8

5x−35

∣∣+ C 228. 1√3ln∣∣x− 7

6 +√

x2 − 72x+ 5

3

∣∣+ C

229. −2√1 + x− x2 − 9 arcsin 2x−1√

5+ C 230. −8

√5 + 2x− x2 − 3 arcsin x−1√

6+ C

231. x√x2 − 2x+ 5− 5 ln

∣∣x− 1 +√x2 − 2x+ 5

∣∣+ C 232.(x−32

)√x2 + 2x+ 2 + 1

2 ln∣∣x+ 1 +

√x2 + 2x+ 2

∣∣+ C

233.(−x2

3 + 56x−

196

)√1− 2x− x2 − 4 arcsin x+1√

2+ C 234.

(x2

3 + 56x+ 1

)√x2 + 4x+ 3− 3

2 ln∣∣x+ 2 +

√x2 + 4x+ 3

∣∣+ C

235. 12 (3− x)

√1− 2x− x2 + 2arcsin x+1√

2+ C 236. x+2

2

√x2 + 4x+ 3− 1

2 ln∣∣x+ 2 +

√x2 + 4x+ 3

∣∣+ C

237. x+22

√x2 + 4x+ 13 + 9

2 ln∣∣x+ 2 +

√x2 + 4x+ 13

∣∣+ C 238. 23√3arcsin 3x+1

2 +(12x+ 1

6

)√1− 2x− 3x2 + C

239. x+12

√3− 2x− x2 + 2arcsin x+1

2 + C 240.(43x

2 − 6x+ 13

)√2 + 3x− x2 + 17

2 arcsin 3−2x√17

+ C

241. − 12 sinx cosx+ x

2 + C 242. 12 sinx cosx+ x

2 + C

243. − 13 sin

2 x cosx− 23 cosx+ C 244. 1

3 sinx cos2 x+ 2

3 sinx+ C

245. 38x−

14 sin 2x+ 1

32 sin 4x+ C 246. 38x+ 1

4 sin 2x+ sin 4x32 + C

247. − cosx+ cos3 x− 35 cos

5 x+ 17 cos

7 x+ C 248. sinx− 23 sin

3 x+ sin5 x5 + C

249. − 13 cos

3 x+ C 250. 115 sin

3 x(3 cos2 x+ 2) + C

251. 115 cos

3 x(3 cos2 x− 5) + C 252. 18 sin

8 x+ C

253. − t3

3 + 2t5

5 −t7

7 + C, kde t = cosx 254. z5

5 −z7

7 + C, kde z = sinx

255. t8

8 −t6

6 + C, kde t = cosx 256. sin 2x(− 18 cos

2 x+ 116 ) +

x8 + C

257. 116x+ 1

32 sin 2x−718 sinx cos

3 x+ 16 sinx cos

5 x+ C 258. − 14 −

14 sin 2x−

116 sin 4x+ 1

24 sin3 2x+ C

259. 1cos x + C 260. cosx+ 1

cos x + C261. ln | cosx|+ 1

2 cos2 x + C 262. 13 cos3 x −

1cos x + C

263. − 14 sin4 x

+ C 264. 13 sin

3 x− 2 sinx− 1sin x + C

265. sin x2 cos2 x + 1

4 ln∣∣ sin x−1sin x+1

∣∣+ C 266. cosx+ 12 ln

∣∣ cos x−1cos x+1

∣∣+ C

267. 13 cos

3 x+ cosx+ 12 ln

∣∣ cos x−1cos x+1

∣∣+ C 268. − sinx+ 12 ln

∣∣ sin x+1sin x−1

∣∣+ C

269. tg x− x+ C 270. tg x+ 12 sinx cosx−

32x+ C

271. 13 arctg

sin x3 + C 272. 1

4 ln∣∣ sin x+1sin x+5

∣∣+ C

273. 12 ln

∣∣ cos x+2cos x

∣∣+ C 274. 1cos x−1 + C

275. 12 cos

2 x− 2 cosx+ 3 ln | cosx+ 2|+ C 276. 12(1−sin x)2 + C

277. sin x2 cos2 x + 1

4 ln∣∣ sin x+1sin x−1

∣∣+ C 278. 12 ln

∣∣ sin x+1sin x−1

∣∣+ C

279. − 1sin x + 1

2 ln∣∣ sin x+1sin x−1

∣∣+ C 280. ln | tg x|+ C

281. 12 arctg

(2 tg x

2

)+ C 282. 1

2−tg x2+ C

283. ln∣∣ tg2 x

2+3

tg2 x2+1

∣∣+ 4√3arctg

(1√3tg x

2

)+ C 284. x tg x

2 + C

285. 2√65

arctg(√

513 tg

x2

)+ C 286. x− tg x

2 + C


287. 14 ln (tg

2 x+ 1)− 12 ln (tg x+ 1) + x

2 + C 288. 41−tg x

2− x+ C

289. ln (tg x2 − 5)− ln (tg x

2 − 3) + C 290. arctg (tg x2 + 1) + C

291. ln | sinx+ cosx|+ C 292. 13 arctg

tg x3 + C

293. 16 arctg (

2 tg x3 ) + C 294. 1

3 arctg(3 tg x) + C

295. 15 ln | tg x− 4| − 1

5 ln | tg x+ 1|+ C 296. ln3√tg x−1

6√

tg2 x tg x+1−√33 arctg 2 tg x+1√

3+ C

297. 15 ln (tg x− 5)− 1

5 ln (tg x) + C 298. 13 tg3 x + C

299. tg x+ tg3 x3 + C 300. −1

3 tg3 x −1

tg x + C

301. −12 sin arctg x√

2

+ C 302. 1a ln

∣∣ arctg xa

2

∣∣+ C

303.√x2 − 9 + 3 arctg 3√

x2−9 + C 304. 12x√x2 − 4 + 2 ln

∣∣x+√x2 − 4

∣∣+ C

305. −√4−x2

x − arcsin x2 + C 306. x√

1−x2+ C

307. x9√9+x2

+ C 308.√x2−99x + C

309. ln (t+1)2

|t| + C, kde t = ex 310. 14 ln

(e2x + 4

)− 1

2x+ 12 arctg

ex

2 + C

311. 2ex + arctg ex + C 312. 18 arctg

e4x

2 + C

313. 1ln a ln ax

ax+1 + C 314. arctg(2t−√3) + arctg(2t+

√3) + C, kde t = ex

315.(ln3 x− 3 ln2 x+ 7 lnx− 7

)x+ C 316. ln

∣∣ 1−√1−x2

x

∣∣− 1x arcsinx+ C

317. ln 1√1−t2 + C, kde t = lnx 318.

√1 + x2 · arctg x− ln

∣∣x+√1 + x2

∣∣+ C

319. 1arccos x + C 320. 4

5

√(1−

√x)5 − 4

3

√(1−

√x)3 + C

321. − ln(e−x +

√e−2x − 1− arcsin ex + C

)322. − 3

43√(1 + 2 cosx)2 + C

323. 16 ln

1+x2

x2 − arctg x3x3 − 1

6x2 + C 324. 18 sin2 x

2− 1

4 ln∣∣ tg x

2

∣∣+ C

325. C −√22 arctg(

√2 cotg 2x) 326. − ln

∣∣ 2+t+2√1+t+t2

2t

∣∣+ C, kde t = ex

327. − 1earcsin x + C 328. 2

3 (arctg ex)

3+ C

329. (ln | arctg x|)22 + C 330. − 2

3

√(arccosx)3 + C


Chapter 4. (Differential equations)

Aim

Introduce the students to the analytical solution of some types of ordinary differential equations and describe the general solution and particular solution.

Objectives

1. Determine the order of a differential equation, specify the number and type of initial conditions.

2. Classify a first order differential equation, learn to calculate some types of such equations.

3. A work of integration constants, determination of integration constants according to the initial conditions.

4. Analyze differential equation with constant coefficients of the second or higher order and understand various methods of their solution.

Prerequisities

function; graph of a function; curve; derivative; integral

Chapter 4. (Differential equations) – 1

Introduction

In the chapter, special kinds of ordinary differential equations are discussed, the solution is search either in general form or in particular form given by aninitial conditions. As long as a lot of physical processes or geometrical dependencies are expressed by differential equations, the equations are frequentlyused in solving civil engineering problems. They can be found in mechanical analysis of constructions or their parts, in heat transfer problems or particletransfer problems in buildings, or in optimizing and controlling of the construction management. Many differential equations do not provide an analyticalsolution in a closed form, nevertheless, in several basic problems such a solution can be found which is the topic of the present chapter.

First, we will see which characteristics are important for taking into account an appropriate method for solving particular equation. Basic methodologiesof first order differential equation solutions cover the separation of variables and variation of constants. The second tool is especially useful for solvinglinear differential equations. In addition to these basic types, also another types of differential equations will be shown which are usually solved by asubstitution to transform a more complicated equation to a simpler, in fact separable or linear.

The general solution will be looked for in all cases, however a special attention will also be paid to determination of particular solution based on prescribedinitial conditions. From the physical point of view or its technical implementation, the solution of a problem with initial conditions is more important.

The next part is focused on the solution of linear differential equation of the second or higher order with constant coefficients. Two commonly usedmethod used for their solution will be discussed. First, a method of special-form right-hand side uses a method of indeterminate coefficients for a seriesof right-hand side functions. Second, if the function fails to be considered as a special form function, the method of constants variation is used as ageneral tool for obtaining linear equations. Also in this case, appropriate initial conditions are required to correctly interpret the results in physical sense.A few words will also be spent on solving the systems of first order differential equations which can be solved by substitutions to come to one higherorder equations.

The final part presents a series of word problems which can be converted to the solution of differential equations. The problems cover geometricimplementations and also physical ones.

Differential equations – Basic notions

Before we start to solve the differential equations, we present some basic differential equations terminology.

The n-th order ordinary differential equation is called the equation of the form

F (x, y, y′, y′′, . . . , y(n)) = 0,

where y=y(x) is an unknown function of one real variable x and F is an arbitrary function of n+2 unknowns. The number n is calld the order of the


differential equation. If the function F is a degree m polynomial with respect to variables y, y′, y′′, . . . , y(n), the number m is called the degree of thedifferential equation. The degree one differential equation is called linear.

The solution or the integral of the differential equation is any function y=ϕ(x) satisfying the relation

F (x, ϕ(x), ϕ′(x), ϕ′′(x), . . . , ϕ(n)(x)) = 0.

The graph of the function is called integral curve of the differential equation.

The integralφ(x, y, c1, . . . , cn) = 0

of the aforementioned differential equation containing n arbitrary constants is called a general integral, or a general solution of the differential equation.The constants c1, . . . , cn are called integration constants. If the integration constants are substituted by particular values, the particular solution isobtained.

Let for the solution y=ϕ(x) of the aforementioned differential equation following conditions hold at x=x0:

y(x0) = b0, y′(x0) = b1, . . . , y(n−1)(x0) = bn−1,

where b0, b1, . . . , bn−1 are arbitrary constants. Then these conditions are called Cauchy initial conditions.

The solution of a differential equations given together with initial conditions is called also a Cauchy problem.

Example 1 Determine the order and the degree of the differential equation

y2 y′′+y sinx y′=x4.

Then specify required number of initial conditions for a unique determination of integration constants in general solution and then choose them arbitrarily.

Solution. The equation can be written, using the function F as follows:

F (x, y, y′, y′′) = y2 y′′+y sinx y′ − x4 = 0.

The differential equation contains the second derivative of the unknown function y as the highest, thus this is the second order equation. It impliesthat there are two integration constants. Moreover, the function F (x, y0, y1, y2)=y

20y2 + sinx y0y1−x4 is with respect to the variables y0=y, y1=y′ and

y2=y′′ the third degree polynomial (due to the term y20y2), hence the degree of the differential equation is three.

We need two initial conditions to uniquely determine integration constants. They could be prescribed, for example as

y(1) = 1, y′(1) = 2.


Differential equations – Basic first-order equations

Rate of change of a physical quantity is expressed by the first derivative so that finding a relation between such a quantity and its rate leads to differentialequations of the first order. We will se the principal methods for their solving.

A general for of a differential equation of the first order is

F (x, y, y′) = F (x, y,dy

dx) = 0,

which is usually expressed in explicit form with respect to the derivative y′=f(x, y). The first step in solving the equation identification of variableseparation or of equation linearity.

If a differential equation can be expressed in the formϕ(y)dy = ψ(x)dx

for some continuous functions ϕ and ψ, the equation is called the equation with separated variables.

The differential equation of the formϕ1(x)ψ1(y)dx = ϕ2(x)ψ2(y)dy

is called the equation with separable variables. Being such an equation divided by the product ϕ2(x)ψ1(y), the equation with separated variables isobtained. The general integral is given by the following expression:∫

ϕ1(x)

ϕ2(x)dx−

∫ψ2(y)

ψ1(y)dy = C.

The solution is often written in an implicit form Φ(x)−Ψ(y)−C=0, but if it is possible we prefer an explicit form y=Υ(x,C).

A linear first-order differential equation with non-zero right-hand side is usually written in the form

y′ + p(x)y = q(x),

where p and q are continuous functions with the only variable x. In the case of q(x)≡0, the equation is called the equation with zero right-hand side(frequently also without right-hand side). The linear differential equation can be solved in different ways:

A. It can be proved that the general solution of a linear differential equation has the form

y =

[∫q(x)e

∫p(x)dxdx+ C

]e−

∫p(x)dx.


B. Applying themethod of variation of constants, the original problem is solved first with zero right-hand side by separating the variables. The solutionof the simplified equation y0 can be always written in the form

y0(x) = ce−∫p(x)dx.

In the second step, it is considered that the constant c is a variable c = C(x) and the solution of the original problem y (non-zero right-hand side)is looked for in the form

y(x) = C(x)e−∫p(x)dx,

such that the unknown function C is determined, if y and its derivative y′ are substituted into the original equation with non-zero right-hand side.

The Cauchy problem prescribes besides the differential equation also the initial condition

y(x0) = b0.

This condition uniquely determines the integration constant C in general integral.

Example 2 Solve the differential equation(1 + ex)yy′ = ex.

Solution. As long as y′=dydx, the equation is expressed as (1 + ex)y dy

dx=ex. The equation is multiplied by the expression dx

1+exto obtain ydy= ex

1+exdx. The

last equation is integrated and the general solution is rendered

y2

2− ln(1 + ex) = C.

Example 3 Find the particular solution of the equation y′ cos2 x=y ln y which satisfies the initial condition y(π4)=e.

Solution. The equation can be written as dydx

cos2 x=y ln y, where the variables are separated to obtain dyy ln y

= dxcos2 x

. The integral of the equationprovides the general solution

ln | ln y| = tg x+ C.

The initial condition gives y=e for x=π4, thus after substitution to the general solution, the integration constant reads C= − 1. The implicit form of

the particular solution isln | ln y| − tg x+ 1 = 0.


Example 4 Solve the equationy′ + 2xy = 2xe−x

2

.

Solution. We use the variation of variables to solve the equation. First, the equation with zero right-hand side is solved

y′ + 2xy = 0,

with subsequent modifications: dyy

=− 2xdx and ln |y|=− x2+ ln c. The general solution of the equation is

y=ce−x2

.

Supposing the variation c=C(x), the derivative of the original solution is

y′ = C ′(x)e−x2 − C(x)2xe−x

2

.

Both y and y′ are substituted into the original equation. The terms with C are eliminated and remains only C ′(x)e−x2=2xe−x

2 , which provides the variedconstant C(x)=x2+c1. The function is substituted for c in general solution of the equation with zero right-hand side to obtain the general solution ofthe equation with non-zero right-hand side

y = (x2 + c1)e−x2 .

Remark 1 If an initial condition, say y(0)=1 is added to the differential equation in Example 4, the integration constant c1 can be determinedby the condition 1=(0+c1)e

−02 . The pertinent particular solution has the form

y = (x2 + 1)e−x2

.

The graphs in Fig. 4.1 show particular solutions of the differential equation for various integration constants c1 for x≥0. The solution correspondingto the given initial condition is drawn by thick red line.

Differential equations – Other types of first-order differential equations

If the differential equation does not belong to any of the basic types, the equation is modified in order to obtain either a linear equation or an equationwith separated variables.


O

c1=12

c1=1c1=

32

c1=2

c1=3

x

y

Figure 4.1: Particular solution for various values of c1.

A function f(x, y) is called degree n homogeneous with respect to variables x and y, if for any t holds

f(tx, ty) = tnf(x, y).

The differential equation y′=f(x, y) is called homogeneous, if f(x, y) is degreee zero homogeneous function, i.e. f(tx, ty)=f(x, y) for all t. It alsoimplies f(x, y)=f(1, y

x) and the homogeneous differential equation can be written in the form

y′ = ϕ(yx

).

The substitution z= yxconverts the homogeneous equation to an equation with separable variables, because

y = zx, y′ = z′x+ z, z′ =dz

dx.

Hence dzdxx+z=ϕ(z), which gives

dz

ϕ(z)− z=

dx

x.

Another type of an equation is the Bernoulli equation. It is an equation of the form

y′ + p(x)y = q(x)yα.

Special case cover the linear equation with non-zero right-hand side for α=0, or the linear equation with zero right-hand side for α=1. The other valuesof α can be treated by the substitution

z = y1−α, z′ = (1− α)y−αy′.


It converts the Bernoulli equation to a linear one. Notice that for α>0 the zero function y≡0 is a particular solution of the equation.

A slightly different type of an equation is the following:M(x, y) +N(x, y)y′ = 0.

Such an equation is called exact differential equation, if it holds

∂M(x, y)

∂y=∂N(x, y)

∂x.

The general solution of the exact differention equation has an implicit form f(x, y)=C, where the function f(x, y) satisfies the conditions

∂f(x, y)

∂x= M(x, y),

∂f(x, y)

∂y= N(x, y).

Example 5 Solve the differential equation(x2 + y2)y′ = 2xy.

Solution. The differential equation is homogeneous as it can be written as

y′ =2xy

x2 + y2

and the right-hand side function satisfies the following relations:

2(tx)(ty)

(tx)2 + (ty)2=

2xyt2

(x2 + y2) t2=

2xy

x2 + y2=

2 yx

1 +(yx

)2 .Put y=xz with y′=z′x+z and substitute it to the modified equation rendering z′x+ z = 2z

1+z2, or equivalently

x(1 + z2)z′ + z3 − z = 0.

This is an equation with separable variables which provides after separation the integral:∫1 + z2

z3 − zdz =

∫1

xdx.


Both sides of the equation are integrated (the left-hand side is decomposed to partial fractions) and the integral reads

ln

∣∣∣∣z2 − 1

z

∣∣∣∣− ln |x| = lnC2,

where C2=eC1>0. Trivial modifications provide the equation x|z2 − 1|=C2|z|. Adding negative values to integration constant, the absolute value canbe removed and the general solution reads simply as

xz2 − x = C,

where C 6=0. Utilizing the back substitution from y=xz, the general solution has the form

y2 − x2 = Cy.

Remark 2 The solution of the Cauchy problem, where the initial condition is y(0)=2, the integration constant C is determined by the condition22 + 00=C · 2. The implicit function of the particular solution is

y2 − x2 = 2y.

Realize that the solution initiates at given point (physically). Thus, talking about the solution for x<0 lose the sense as these xs do not belongto the domain of the solution. Let us try to demonstrate it by the graph in Fig. 4.2. The general solution is a set of all functions with variousconstants C, in the figure e.g. C∈{1

2, 23, 2, 3}. However, our particular solution (the thick red line) is only a part of a curve beginning at the point

[0, 1].

Example 6 Solve the equation

y′ +2

xy = x4y2.

Solution. It is the Bernoulli differential equation with α=2. As 2>0 one particular solution is y=0. Let us multiply the equation by the term y−2. Weobtain

y−2y′ +2

xy−1 = x4.

Now, put z=y−1 with z′=− y−2y′ and substitute it to the original equation rendering

−z′ + 2

xz = x4.


O

C=12

C=1C=3

2

C=2

C=3

C=12

C=32

C=2C=3

12

1

32

2

3

x

y

Figure 4.2: Particular solutions for various C.

This is a linear differential equation, whose solution

z = −x5

3+ C1x

2

can be found e.g. by the method of variation of variable as used in Example 4. Using the back substitution for z, the solution of the Bernoulli differentialequation reads

y =3

Cx2 − x5, y = 0,

where C=3C1.

Remark 3 The zero solution does not correspond to any value C, though it can be obtained by a limit with C→∞. For fixed x, the limit isalways zero:

y(x) = limC→∞

3

Cx2 − x5= 0.


Example 7 Solve the equation(3x2 + 6xy2) + (6x2y + 4y3)y′ = 0.

Solution. First, let us show that the equation is an exact differential equation. The partial derivatives of the terms M (without y′) and N (with y′) withrespect to pertinent variables read

∂M(x, y)

∂y=∂(3x2 + 6xy2)

∂y= 12y,

∂N(x, y)

∂x=∂(6x2y + 4y3)

∂x= 12y.

As the derivatives are equal to each other, the equation is exact and its integral f satisfies the relation

∂f(x, y)

∂x= M(x, y) = 3x2 + 6xy2.

It implies

f(x, y) =

∫(3x2 + 6xy2)dx = x3 + 3x2y2 + ϕ(y),

∂f(x, y)

∂y= 6x2y + ϕ′(y).

Simultaneously, the other condition should be valid ∂f(x,y)∂y

=N(x, y)=6x2y + 4y3, thus 6x2y+ϕ′(y)=6x2y+4y3. The last relation provides ϕ(y)=y4.The general solution of the given equation is any function

x3 + 3x2y2 + y4 = C

with an arbitrary integration constant C.

Differential equations – Linear differential equations of the second or higher order with constant coefficients

The equations of a higher order cannot be easily integrated in general case, nevertheless a linear equation with constant coefficients can be solve by asimilar scheme as the linear equations of the first order.

A linear equation of any order n always enables to obtain the solution in two steps. In the first step, an equation with zero right-hand side is solved. Inthe second step a particular solution of the equation with non-zero right-hand side is found. The method of finding this particular solution depends onthe function which appears in the right-hand side of the equation. The general solution of a linear equation is a combination of the two components.As an additional final step, if it is required by the problem definition, the integration constants are determined satisfying the initial conditions

y(x0) = b0, y′(x0) = b1, . . . , y(n−1)(x0) = bn−1.


An equationy(n) + a1y

(n−1) + · · ·+ an−1y′ + any = 0,

where ai, i=1, 2, . . . , n are real numbers, is called the linear differential equation with constant coefficients with zero right-hand side. The polynomialequation with the unknown r and with the same coefficients of the form

rn + a1rn−1 + a2r

n−2 + · · ·+ an−1r + an = 0

is called the characteristic equation of the differential equation. The general solution of the equation with zero right-hand side has a form

y = c1y1 + c2y2 + · · ·+ cnyn,

where y1, y2, . . . , yn are linearly independent solutions of the equation and c1, c2, . . . , cn are arbitrary (integration) constants.

If r1 is a multiple root of the characteristic equation with multiplicity k, then the functions er1x, xer1x, x2er1x, . . . , xk−1er1x are linearly independentsolutions of the equation with zero right-hand side.

If α + βi is a k-multiple imaginary root of the characteristic equation, then the functions

eαx cos βx, xeαx cos βx, x2eαx cos βx, . . . , xk−1eαx cos βx,

eαx sin βx, xeαx sin βx, x2eαx sin βx, . . . , xk−1eαx sin βx,

are linearly independent solutions of the equation with zero right-hand side.

An equationy(n) + a1y

(n−1) + · · ·+ an−1y′ + any = f(x),

is called linear differential equation with constant coefficients with non-zero right-hand side. Its general solution is obtained as a sum of the generalsolution of the equation with zero right-hand side and an arbitrary particular solution of the equation with non-zero right-hand side. If the right-handside f(x) is given in any of the forms below, the equation is said to have special right-hand side. In these cases, the form of particular solution can beguessed and unknown parameters in the form are calculated by the method of indeterminate coeficients. The method is used under following conditions,for example:

A. f(x)=Pm(x)eαx, where Pm(x) is an m-degree polynomial and α is a real number: the particular solution can be written in the form

y = (b0 + b1x+ b2x2 + · · ·+ bmx

m)eαxxk,

if α is a k-multiple root of the characteristic equation (if α is not a root of the characteristic equation, k=0). The real numbers b0, b1, . . . , bm arethe unknown parameters calculated by the method of indeterminate coefficients.


B. f(x)=eαx(A cos βx+B sin βx), where α, β are real numbers: the particular solution can be written in the form

y = eαx(M cos βx+N sin βx)xk,

if α+βi is a k-multiple root of the characteristic equation. The constants M and N are again calculated by the method of indeterminatecoefficients.

Another possibility for finding the solution of the equation with non-zero right-hand side is the Lagrange method of variation of constants. This methodis usually used only in the case, if the right-hand side cannot be expressed as a special form function or a sum of such functions. Let y1, y2, . . . , yn aren linearly independent solutions of the equation with zero right-hand side. The general solution of the equation with non-zero right-hand side is of theform

y = C1(x)y1 + C2(x)y2 + · · ·+ Cn(x)yn,

where the functions C1(x), C2(x), . . . , Cn(x) are obtained as a solution of the following equation system:

C ′1(x)y1 +C ′2(x)y2 + · · ·+C ′n(x)yn =0,

C ′1(x)y′1 +C ′2(x)y′2 + · · ·+C ′n(x)y′n =0,

C ′1(x)y′′1 +C ′2(x)y′′2 + · · ·+C ′n(x)y′′n =0,

......

C ′1(x)y(n−1)1 +C ′2(x)y

(n−1)2 + · · ·+C ′n(x)y(n−1)n =f(x).

The determinant of the left-hand side matrix in the system is call the Wronski determinant or Wronskian W (C1, C2, . . . , Cn). It can be proved thatthe functions f1, f2, . . . , fk are linearly dependent in the interval 〈a, b〉, if W (f1, f2, . . . , fk)=0 for any x∈〈a, b〉. If W (f1, f2, . . . , fk) 6=0 at least at onepoint x∈〈a, b〉, the functions are linearly independent. This implies that the system can be solved by Cramer rule and the solution is written in the form

y = y1

∫W1(x)

W (x)dx+ y2

∫W2(x)

W (x)dx+ · · ·+ yn

∫Wn(x)

W (x)dx,

were W (x), W1(x), W2(x), . . . ,Wn(x) are pertinent determinants necessary for solving the system by the Cramer rule.

Finally, recall that linearity of the equation enable to use the superposition principle to find the general solution. If the solution of the equationwith zero right-hand side is denoted by y0, the right hand side is decomposed to a sum of s function (say, in special form, but not necessarily)f(x)=f1(x)+f2(x)+ · · ·+fs(x) and the solutions of the equation with the right-hand side fi(x) are respectively denoted as yi, then the solution reads

y = y0 + y1 + y2 + · · ·+ ys.


Example 8 Solve the differential equationy′′′ − 2y′′ − 3y′ = 0.

Solution. An equation with zero right-hand side is solved, we need to find the characteristic equation. It reads

r3 − 2r2 − 3r = 0.

The roots are r1=0, r2=−1, r3=3. All the roots are simple and real, thus three independent solutions of the differential equation are: e0x, e−x and e3x.The general solution then reads

y = c1 + c2e−x + c3e

3x.

Example 9 Solve the differential equationy′′′ + 2y′′ + 3y′ = 0.

Solution. An equation with zero right-hand side is solved again. The characteristic equation reads.

r3 + 2r2 + r = 0.

The roots are r1=0 and r2=r3=−1. As the root “−1” is double, the pertinent paier of linearly independent solutions is e−x and xe−x. The third linearlyindependent solution is a constant function so that the general equation of the differential equation is

y = c1 + c2e−x + c3xe−x.

Example 10 Solve the differential equationy′′ + 2y′ + 5y = 0

with initial conditions y(0)=2, y′(0)=0.

Solution. The characteristic equation isr2 + 2r + 5 = 0.


The roots of the equation are imaginary (negative discriminant of the quadratic trinomial): r1=−1+2i and r2=−1−2i. The functions e−x cos 2x ande−x sin 2x are linearly independent solutions of the equation and the general forma reads

y = c1e−x cos 2x+ c2e

−x sin 2x.

The integration constants can be determined by the initial conditions which also require the derivative of the general solution:

y′ = c1(−e−x cos 2x− 2e−x sin 2x

)+ c2

(−e−x sin 2x+ 2e−x cos 2x

).

The given initial conditions render

2 = c1e−0 cos (2·0) + c2e

−0 sin (2·0) = c1,

0 = c1(−e−0 cos (2·0)− 2e−0 sin (2·0)

)+ c2

(−e−0 sin (2·0) + 2e−0 cos (2·0)

)= −c1 + 2c2.

The solution of the two equation system is a couple c1=2, c2=1 which provides the solution of the Cauchy problem in the form

y = e−x (2 cos 2x+ sin 2x) .

Example 11 Solve the differential equationy′′ + y′ = 4x2ex.

Solution. The root of the characteristic equation r2+r=0 are r1=0, r2=−1, thus the general solution of the equation with zero right-hand side is

y0 = c1 + c2e−x.

As α=1 is not a root of the characteristic equation, the particular solution y∗ of the equation with non-zero right-hand side can be guessed in the form

y∗ = (Ax2 +Bx+ C)ex.

This function, for appropriately chosen parameters A, B, C is a solution of the equation, so it can be substituted into the equation together with itsderivatives

y∗′ = (2Ax+B)ex + (Ax2 +Bx+ C)ex,

y∗′′ = 2(2Ax+B)ex + 2Aex + (Ax2 +Bx+ C)ex.


The substitution renders2Ax2 + (6A+ 2B)x+ 2A+ 3B + 2C = 4x2.

Comparing the coefficients by the same power of x, we have

2A = 4,

6A+ 2B = 0,

2A+ 3B + 2C = 0,

satisfied for A=2, B=−6, C=7. The particular solution is then

y∗ = (2x2 − 6x+ 7)ex

and the general solution of the differential equation is a sum of y0 and y∗

y = c1 + c2e−x + (2x2 − 6x+ 7)ex.

Example 12 Solve the differential equationy′′ + 10y′ + 25y = 4e−5x


Solution. The characteristic equation r2+10r+25=0 has a double root r1=r2=−5, hence the general solution of the equation with zero right-hand sideis

y0 = c1e−5x + c2xe−5x.

Moreover, as α=−5 is a root of characteristic equation with multiplicity k=2, the particular solution of the equation with non-zero right-hand side canbe expressed as

y∗ = Ae−5xx2.

The solution y∗ and its derivatives

y∗′ = 2Axe−5x − 5Ax2e−5x,

y∗′′ = 2Ae−5x − 20Axe−5x + 25Ax2e−5x.


are again substituted to the equation providing2Ae−5x = 4e−5x,

with the solution A=2. The particular solution readsy∗ = 2x2e−5x

and the general solution is obtained by adding the solution y0

y = y0 + y∗ = c1e−5x + c2xe−5x + 2x2e−5x.

Now, the integration constants should be determined. To use the initial conditions, the solution derivative is required

y′ = −5c1e−5x + c2

(e−5x − 5xe−5x

)+ 2

(2xe−5x − 5x2e−5x

).

The start point provides the relations

0 = c1e−5·0 + c2 · 0e−5·0 + 2 · 02e−5·0 = c1,

1 = −5c1e−5·0 + c2

(e−5·0 − 5 · 0e−5·0

)+ 2

(2 · 0e−5·0 − 5 · 02e−5·0

)= −5c1 + c2,

whose solution is a constant couple c1=0, c2=1. The Cauchy problem solution then reads

y = xe−5x (1 + 2x) .

Example 13 Solve the differential equationy′′ − 6y′ + 9y = 25ex sinx.

Solution. Also in this case the characteristic equation r2−6r+9=0 has a double root r1=r2=3, the solution y0 (zero right-hand side) can be expressedas

y0 = c1e3x + c2xe3x.

As to the special right-hand side eαx sin βx=ex sinx pertains the number α+βi=1+i (or α−βi=1−i), which is not the root of the characteristic equation,the particular solution can be guessed in the form

y∗ = ex(M cosx+N sinx).


The derivatives of this functions are

y∗′ = ex(M cosx+N sinx) + ex(−M sinx+N cosx),

y∗′′ = ex(M cosx+N sinx) + 2ex(−M sinx+N cosx) + ex(−M cosx−N sinx)

an their substitution to the given equation renders

(3M − 4N) cosx+ (4M + 3N) sinx = 25 sin x.

Comparing the coefficients by the functions cosx and sinx, the system of linear equations is obtained

3M − 4N = 0,

4M + 3N = 25,

whose solution is M=4, N=3. The general solution of the differential equation with non-zero right-hand side is

y = c1e3x + c2xe3x + ex(4 cosx+ 3 sinx).

Example 14 Solve the differential equationy′′ + y′ = 5x+ 2ex


Solution. The roots of the characteristic equation r2+r=0 are r1=0, r2=−1 which provides the solution y0 in the form

y0 = c1 + c2e−x.

The particular solution of the equation with non-zero right-hand side can be obtained by the superposition principle as a sum of particular solutions y1and y2 of the equations

y′′ + y′ = 5x,

andy′′ + y′ = 2ex,

respectively. The form of the particular solution y1 is y1=(Ax+B)x, the form of the particular solution y2 is y2=Cex. Both solutions are differentiatedand substituted into pertinent equations. The indeterminate coefficients are calculated: A=5

2, B=−5, C=1. The general solutio of the original equation

isy = c1 + c2e

−x +5

2x2 − 5x+ ex.


Finally, the integration constants will be determined. With

y′ = −c2e−x + 5x− 5 + ex,

the initial conditions provide the constants c1 and c2 by the relations

0 = c1 + c2e−0 +

5

202 − 5 · 0 + e0 = c1 + c2 + 1,

0 = −c2e−0 + 5 · 0− 5 + e0 = −c2 − 4.

The solution of the system c1=3, c2=−4 renders the solution of the Cauchy problem

y =5

2x2 − 5x+ 3− 4e−x + ex.

Remark 4 Let us draw some particular solutions for various values of integration constants. The graphs in Fig. 4.3 present five such couples,including the solution of the Cauchy problem in Example 14. This solution is drawn by the thick red line and naturally only from the point of

O

c1=1, c2=0

c1=3, c2=−4

c1=0, c2=1

c1=4, c2=−1

c1=1, c2=−2 x3

y=3y=x

y

Figure 4.3: Particular solutions for various values of integration constants.

the initial conditions: the graph begins at the point O and its tangent is the line y=x which means unit derivative at this point. Let us noticethat the scales of the coordinate axes are different due to exponential function which increases (or decreases) too fast for x just a bit greater thanone.


The graphs also show that while for the negative x the particular solutions are rather distinct, for the positive x are similar. This is again causedby the exponential function ex present in the particular solution. All other function are for sufficiently large x negligible with respect to thisexponential.

Example 15 Find the general solution of the equation

y′′ + y =1

cosx.

Solution. The roots of characteristic equation r2+1=0 are r1=i, r2=−i, hence the solution with zero right-hand side is

y0 = c1 cosx+ c2 sinx.

We put c1 = C1(x) and c2 = C2(x) and use the method of variation of constants in the next calculation. We obtain the following system of equationsfor unknown functions C1, C2:

C ′1(x) cosx+ C ′2(x) sinx = 0,

−C ′1(x) sinx+ C ′2(x) cosx =1

cosx,

whose solution is C ′1(x) = − tg x a C ′2(x) = 1. The functions C1, C2 are rendered by integration

C1(x) = ln | cosx|+ c3, C2(x) = x+ c4.

The general solution can be written by the expression

y = (ln | cosx|+ c3) cosx+ (x+ c4) sinx.

Remark 5 Let us notice the last form of the solution. As the solution of a linear equation, it can be expressed as a sum of the solutions withzero right-hand side and with non-zero right-hand side. Let us rewrite it in the form

y = (c3 cosx+ c4 sinx) + (ln | cosx| cosx+ x sinx) .

The term in the first parentheses is y0, though the notation for integration constants is different then above. In the second parentheses, theparticular solution is written.


Differential equations – systems of first-order differential equations

Relations of various physical quantities are frequently more complicated and require to solve not only one equation by a whole system of differentialequations.

A set of first-order differential equationsx′1 = f1(t, x1, x2, . . . , xn),

x′2 = f2(t, x1, x2, . . . , xn),

...x′n = fn(t, x1, x2, . . . , xn),

is call a system of differential equations in normal form, the number n is called the order of the system. The solution of the system is considered anyn-component vector of differentiable functions (x1(t), x2(t), . . ., xn(t)) such that substituting it to the system, n equalities are obtained.

The system can be solved e.g. by the elimination method which converts the system of n equations of the first order to one equation of the order nwith one unknown function, say x1. The form of the resulting equation is

x(n)1 = φ(t, x1, x

′1, . . . , x

(n−1)1 )

with the solution x1=ϕ1(t, c1, c2, . . ., cn). Subsequently, other unknown function are calculated until all n components of the solution are rendered

x1 = ϕ1(t, c1, c2, . . . , cn),

x2 = ϕ2(t, c1, c2, . . . , cn),

...xn = ϕn(t, c1, c2, . . . , cn).

Remark 6 Physically, the variable t is often interpreted as time and pertinent time-derivative distinguished by a “dot” instead of a “dash”. This israther useful if we tend to differentiate with respect to x which in the present case is the vector of unknown functions to be found. Moreover, the“dash” can be used in more complicated situations for differentiation with respect to space variables in time-space partial differential equations.Thus, in what follows, the “dot” notation will be used.

A system of two equations can be written in normal form as follows:

x1 = f1(t, x1, x2),

x2 = f2(t, x1, x2).


If the integration constants should be determined uniquely, two initial conditions have to be prescribed, e.g.

x1(0) = b0, x2(0) = b2.

Example 16 Find the particular solution of the systemx = 3x+ 8y,

y = −x− 3y,

which satisfies the initial conditions x(0)=6, y(0)=−2.

Solution. First, we find the general solution of the system. The first equation renders

y =1

8x− 3

8x.

Let us differentiate it (with respect to t)

y =1

8x− 3

8x

and let us substitute for y and y into the second equation of the system. Small formatting provides a simple second-order differential equation

x− x = 0

with the general solution x=c1et+c2e

−t. The dependence of y on x provides the expression for the other component of the solution y=−14c1e

t−12c2e−t.

The initial conditions x(0)=6, y(0)=− 2 are used to determine the values of integration constants

6 = c1 + c2,

−2 = −1

4c1 −

1

2c2.

The solution of the algebraic system is c1=4 and c2=2. The particular solution to have been found is thus

x = 4et + 2e−t,

y = −et − e−t.


O

t=0t=0.6

t=1.1t=1.6

t=2.1

t=2.6

x6

−2

yy=−1

4x

Figure 4.4: The solution of the differential equations system with initial conditions.

Remark 7 The found particular solution is draw in Fig. 4.4 by the red line. It can be considered as a graph of a function given in the parametricform, where [x; y] is a point lying on the graph of the function y=y(x) an t is the parameter. Physically, it corresponds to a motion of a masspoint in a plane with respect to time t.

Although, the graph seems almost linear, definitely, it is not a straight line but a part of a curve, a small curvature can be observed in theneighborhood of the initial point. The reason for the shape is that the exponential function e−t is for larger t negligible with respect to the otherexponential function et and for sufficiently large x (and t, too), there holds y .=−1

4x (the blue line in the picture). Moreover, the “motion” is

evidently accelerated: the distance of time-equally-spaced points is increasing.

Differential equations – Word problems

A differential relation is frequently sought in the physical or geometrical formulation of the problem, the solution of a differential equation is then onlyone step leading to the complete solution of the problem.

Example 17 Find a curve passing through the point A[2, 3] for which the segment of any tangent bounded by the intersection points of the tangentwith the coordinate axes is halved by the tangent point.

Solution. Let us choose on a curve, considered as a graph of a function f given as y=f(x), a point B[xB, yB] as it is shown in Fig. 4.5. The equationof the tangent t to the curve f at the point B reads

t : y − yB = f ′(xB) (x− xB) .


O x2

3

y=f(x)t

xB

yB

A

B

Bx

By

y

Figure 4.5: The segment of the tangent is halved by the tangent point.

The tangent intersects the coordinate axes at points Bx a By, whose coordinates are

Bx ∈ t ∩ ox =

{[xB −

yBf ′(xB)

; 0

]},

By ∈ t ∩ oy ={[

0; yB − f ′(xB)xB]}.

As long as the point B lies on t at the midpoint between these intersection points, its coordinates are given by the following relations:

xB =1

2

(yB −

yBf ′(xB)

),

yB =1

2

[yB − f ′(xB)xB

].

The points Bx, By and B lie on one line (the tangent), thus the equations are after some modifications, the same. Let us consider the second equationand ignore the subscript index B as the equation should be valid for any point of f . Writing y′ instead of f ′(x), a linear first-order differential equationwith an initial condition determined by the point A is obtained

y + xy′ = 0, y(2) = 3.

The solution of this Cauchy problem is y= 6xwhich is the equation of the sought curve (a hyperbola).

Example 18 A mass point M of total mass m is moving rectilinearly affected by a constant force G. Simultaneously, due to resistance of thesurroundings there appears the force Fv directly proportional to the velocity of the motion and oriented against the motion direction, the coefficient of


proportionality is β. Moreover, the point is constrained such that the reaction force R is proportional (with the coefficient k) to the deflection fromthe equilibrium position O and oriented to this equilibrium position, see Fig. 4.6. Find the equation of motion x=x(t), if in the beginning the point is

O HMh

x

x(t)

x(t)

G

Fv=β x

R=k x

Figure 4.6: Forces acting on the mass point M .

moved from the equilibrium position O to the distance h (the point H) and then released.

Solution. This classical example from mechanics can be interpreted as a motion of a mass point fixed by a spring of stiffness k which is affected bygravity G and resistance of air Fv. The equation of motion for unknown deflexion x is written in differential form

mx = G−R− Fv,

where G=mg, R=k x, Fv=β x and g is the acceleration due tu gravity. The initial condition are given as follows:

x(0) = h, x(0) = 0.

The equation can be rewritten in the formmx+ βx+ kx = mg.

The characteristic equation mr2+β r+k=0 may have real or imaginary roots, depending on the discriminant D=β2−4mk of the equation:

A. Ak D>0, tak

r1 =−β +

√β2 − 4mk

2m, r2 =

−β −√β2 − 4mk

2m.

B. Ak D<0, tak

r1 =−β + i

√4mk − β2

2m, r2 =

−β − i√

4mk − β2

2m.

C. Ak D=0, tak

r1 = r2 = − β

2m


The particular solution x∗ of the equation with non-zero right-hand side is simple as long as the right-hand side is only a constant. It is also constant:x∗=mg

k. The general solution depends on D and there are three choices for its expression:

A. If D>0, then

x(t) = c1e−β+√β2−4mk2m

t + c2e−β−√β2−4mk2m

t +mg

k.

B. If D<0, then

x(t) = e−β2m

t

[c1 cos

√4mk − β2

2mt+ c2 sin

√4mk − β2

2mt

]+mg

k.

C. If D=0, thenx(t) = e−

β2m

t (c1 + c2t) +mg

k.

The derivatives of the solutions are necessary for calculation of the integration constants. The derivatives also represent the velocity of the motion. Weobtain

A. If D>0, then

x(t) = c1−β +

√β2 − 4mk

2me−β+√β2−4mk2m

t + c2−β −

√β2 − 4mk

2me−β−√β2−4mk2m

t.

B. If D<0, then

x(t) = − β

2me−

β2m

t

[c1 cos

√4mk − β2

2mt+ c2 sin

√4mk − β2

2mt

]+

√4mk − β2

2me−

β2m

t

[−c1 sin

√4mk − β2

2mt+ c2 cos

√4mk − β2

2mt

].

C. If D=0, then

x(t) = e−β2m

t

[− β

2m(c1 + c2t) + c2

].

The initial conditions also provide three different algebraic systems for determination of integration constants:


A. If D>0, then

c1 + c2 = h− mg

k,

−β +√β2 − 4mk

2mc1 +

−β −√β2 − 4mk

2mc2 = 0.

B. If D<0, then

c1 = h− mg

k,

−β2m

c1 +

√4mk − β2

2mc2 = 0.

C. If D=0, then

c1 = h− mg

k,

−β2m

c1 + c2 = 0.

The substitution of the algebraic system solution into the general solution renders

A. If D>0, then

x(t) =

e−β+√β2−4mk2m

t −

(−β +

√β2 − 4mk

)24mk

e−β−√β2−4mk2m

t

(h− mg

k

)+mg

k.

B. If D<0, then

x(t) = e−β2m

t

[cos

√4mk − β2

2mt+

β√4mk − β2

sin

√4mk − β2

2mt

](h− mg

k

)+mg

k.

C. If D=0, then

x(t) = e−β2m

t

(1 +

β

2mt

)(h− mg

k

)+mg

k.


The physical meaning of these options is left up to physicists.

Self-assessment questions and problems for solving

1. Why the number of constants in general solution of a differential equation is equal to the order of the equation?

2. Can you prove the formulas for solving linear differential equations of the first order?

3. Find a reason, why the linear equation with zero right-hand side can be solved by the characteristic equation!

4. Can you prove the formula for solving the higher order differential equation by variation of constants?

5. Boundary conditions are prescribed in the case when the equation is solved on a bounded interval. How would you define the boundary conditionsfor a differential equation of the second order? Is the solution of the differential equation with boundary conditions always unique?

6. Try to find situations in the subjects of your specialization, when the boundary conditions are prescribed instead of initial ones!

7. Referring to your knowledge from other fields, classify and try to solve differential equations used in specialized subjects!

In the problems 1. to 18. solve the differential equations by the method of variable separation.

1. 2y′√x = y . 2. y

′√

1− x2 − y2 − 1 = 0 .

3. x+ xy + y′(y + xy) = 0 . 4. ey(1 + x2)dy − 2x(1 + ey)dx = 0 .

5. ex sin3 y + (1 + e2x) cos yy′ = 0 . 6. x√

1 + y2 + yy′√

1 + x2 = 0 .

7. y′ = 1−2x

y2. 8. xy

′ = (1 + y2) arctg y .

9.1√x

+ y′√y

= 0 . 10. x2(y + 1) + (x3 − 1)(y − 1)y′ = 0 .

11. cos√xdx−

√xdy = 0 . 12. (y + 4)dx+ (x− 5)dy = 0 .

13. y′ = 10x+y . 14. xyy

′ = 1− x2 .


15. y − y2 + xy′ = 0 . 16. 1 + y2 + xyy′ = 0 .

17. 2y − x3y′ = 0 . 18. y′ − xy2 − y2 − xy − y = 0 .

In the problems 19. to 28. find the solution of the differential equation, which satisfies the initial condition.

19.x

1+y− yy′

1+x= 0, y(0) = 1 . 20.

1+y2

1+x2− y′ = 0, y(0) = 1 .

21. y ln y + xy′ = 0, y(1) = 1 . 22. y′ sinx = y ln y, y(π

2) = 1 .

23. 2y′√x = y, y(4) = 1 . 24. sin y cosxy′ = cos y sinx, y(0) = π

4.

25. (1 + ex)yy′ = ex, y(0) = 1 . 26. y′ sinx− y cosx = 0, y(π

2) = 1 .

27. y′√

1− x2 = xy, y(0) = 1e. 28.

12y+1

dy − cotg xdx = 0, y(π4) = 1

2.

In the problems 29. to 46. solve the first order linear differential equations.

29. y′ + 2y = e−x . 30. y

′ − 3xy = x .

31. y′ + 2xy = e−x

2

x. 32. xy

′ + y = lnx+ 1 .

33. y′ − 2

x+1y = (x+ 1)3 . 34. y

′ − yx

= x .

35. y′ cosx− y sinx = sin 2x . 36. y

′ cosx+ y sinx = 1 .

37. y′ + 2y = e2x . 38. xy

′ + y = x3 .

39. xy′ − 2y = x3 cosx . 40. (1 + x2)y′ + y = arctg x .


41. tg xy′ − y = 2 . 42. y′ − 2x

x2+1y = x2 + 1 .

43. y′ − y tg x = 2 cos2 x . 44. y

′ + xx2+1

y = sinx√x2+1

.

45. y′ + 1

x+1y = sinx . 46. y

′x lnx− 2y = lnx .

In the problems 47. to 52. find the solution of the first order linear differential equations satisfying the initial conditions.

47. y′ + 3y = x, y(1

3) = 1 . 48. xy

′ + y = 3x2, y(1) = 1 .

49. y′ + 3

xy = 2

x3, y(1) = 1 . 50. y

′ − y2(x+1)

= ex√x+ 1, y(0) = 0 .

51. y′ + y cotg x = sinx, y(π

2) = 1 . 52. y

′ − y1−x2 = x+ 1, y(0) = 0 .

In the problems 53. to 67. solve the homogeneous differential equations.

53. xy′ = y + x cos2 y

x. 54. y

2 + y′(x2 − xy) = 0 . 55. xy′ − y = xe

yx .

56. xy′ = y ln y

x. 57. y′ =

y2

x2− 2 . 58. x

2 + y2 = 2xyy′ .

59. xy′ cos yx

= y cos yx− x .

60. xy′ − y =

√x2 + y2 . 61. y

′ = 2x+yx

. 62. x3y′ = y(y2 + x2) .

63. y′ = x−y

x−2y . 64. y′ = 2xy

x2−y2 . 65. y′ − y

x= tg y

x.

66. y′ = x

y+ y

x. 67. (x+ y)y′ − y = 0 .

In the problems 68. to 73. find the solution of the first order homogeneous differential equations satisfying the initial conditions.


68. y′ = xy

x2+y2, y(0) = 1 . 69. y +

√x2 + y2 − xy′ = 0, y(1) = 0 .

70. x+ 2y − xy′ = 0, y(2) = 2 . 71. (xy′ − y) arctg yx

= x, y(1) = 0 .

72. (y2 − x2)y′ + xy = 0, y(1) = 1 .

73. x2 + 2xy − y2 = (x2 − 2xy − y2)y′, y(2) = 2 .

In the problems 74. to 85. find the solution of the Bernoulli differential equations.

74. y′ + 2xy = 2xy2 . 75. y

′ + xy = xy3 .

76. y′ + y

x= y2 lnx . 77. y

′ + y = x√y .

78. 2y′ + y = xy. 79. 2y′ lnx+ y

x= cosx

y.

80. y′ + 2

xy = 2 sinx

x

√y . 81. y

′2 sinx+ y cosx = y3 sin2 x .

82. y′ + y

2= 1

2(x− 1)y3 . 83. y

′ − 43y = x+1

3y2.

84. y′ − y

x= 1

2y. 85. xy

′ + y = y2 lnx .

In the problems 86. to 89. find the solution of the Bernoulli differential equations satisfying the initial conditions.

86. y′ + y = 2y2, y(0) = 2 . 87. 2y′ + x

1−x2y = xy, y(0) = 1 .

88. y′ + y

x= y2, y(1) = 1 . 89. y

′ + 2y tg x =√y cos2 x, y(π

4) = 0 .

In the problems 90. to 99. find the general solution of the exact differential equations.


90. (3x2 − 2x− y)dx+ (2y − x+ 3y2)dy = 0 .

91. (6xy + x2 + 3)dy + (3y2 + 2xy + 2x)dx = 0 .

92. (1 + y2

x2)dx− 2 y

xdy = 0 . 93. x(2x2 + y2)dx+ y(x2 + 2y2)dy = 0 .

94. (2x+ x2+y2

x2y)dx− x2+y2

xy2dy = 0 . 95. (1 + e

xy )dx+ e

xy (1− x

y)dy = 0 .

96.x

x2+y2dy − ( y

x2+y2− 1)dx = 0 . 97. ex(x2 + y2 + 2x)dx+ 2yexdy = 0 .

98. x(x+ 2y)dx+ (x2 − y2)dy = 0 . 99. eydx+ (xey − 2y)dy = 0 .

In the problems 100. to 119. solve the differential equations with zero right-hand side.

100. y′′ − 5y′ = 0 . 101. y

′′ − 9y = 0 . 102. y′′ + 3y′ − 4y = 0 .

103. y′′ + 6y′ + 9y = 0 . 104. y

′′ − 4y′ + 13y = 0 . 105. y′′ + 16y = 0 .

106. y′′ + 2y′ + 5y = 0 . 107. y

′′ − 6y′ + 8y = 0 . 108. y′′ + 25y = 0 .

109. y = c1e52x + c2xe

52x0 . 110. 2y′′ + 11y′ − 6y = 0 . 111. 9y′′ + 18y′ + 13y = 0 .

112. y′′ + y = 0 . 113. y

′′′ − y′ = 0 . 114. y′′′ − y′′ = 0 .

115. y′′′ − 3y′′ + 3y′ − y = 0 . 116. y

′′′ − 3y′ − 2y = 0 . 117. y′′′ − 4y′ = 0 .

118. y′′′ + 8y = 0 . 119. y

′′′′ + 4y = 0 .

In the problems 120. to 127. solve the linear differential equations with special right-hand side by the method of indeterminate coefficients.


120. y′′ + 3y′ = ex . 121. y

′′ − 8y′ + 16y = xe2x . 122. y′′ − 2y′ + 3y = x+ 1 .

123. y′′ + y = 4xex . 124. y

′′ − 4y = 10e3x . 125. y′′ − 3y′ + 2y = ex .

126. y′′ + 4y′ + 4y = e−2x .

127. y′′ + y = 2 sin x− cosx .

In the problems 128. to 133. solve the equation y′′−7y′+10y=f(x) for given f(x).

128. 40 . 129. 20x2 − 28x+ 14 . 130. −12e3x .

131. −e2x(6x+ 7) . 132. (9x2 + 6x− 3)e5x . 133. 8e2x sinx .

In the problems 134. to 136. solve the equation 3y′′−4y′=f(x) for given f(x).

134. −32x3 + 84x+ 50 . 135. 4e4x3 . 136. 25 sinx .

In the problems 137. to 139. solve the equation 9y′′−6y′+y=f(x) for given f(x).

137. 4e−x3 . 138. sin x

3. 139. 9x2 − 6x+ 1 .

In the problems 140. to 142. solve the equation y′′+4y=f(x) for given f(x).

140. cos 2x . 141. cos 3x . 142. e−2x .


In the problems 143. to 145. solve the equation y′′−4y′+5y=f(x) for given f(x).

143. e2x . 144. sinx . 145. e2x sinx .

In the problems 146. to 151. solve differential equations.

146. y′′ − y′ − 2y = 4x− 2ex . 147. y

′′ − 3y′ = 18x− 10 cosx .

148. y′′ − 2y′ + y = 2 + ex sinx . 149. y

′′ + 2y′ + 2y = (5x+ 4)ex + e−x .

150. y′′ − 4y′ + 5y = 1 + 8 cosx+ e2x . 151. y

′′ + 5y′ + 6y = e−x + e−2x .

In the problems 152. to 157. find the solution of differential equations satisfying the initial conditions.

152. y′′ − 6y′ + 9y = 9x2 − 12x+ 2, y(0) = 1, y′(0) = 3 .

153. y′′ + y = 2 cos x, y(0) = 1, y′(0) = 0 .

154. 4y′′ + y = 0, y(π) = 2, y′(π) = 3 .

155. y′′ − y′ = −5e−x(sinx+ cosx), y(0) = −4, y′(0) = 5 .

156. y′′′ − y′ = −2x, y(0) = 0, y′(0) = 1, y′′(0) = 2 .

157. y′′ − 4y′ + 3y = e2x, y(0) = 0, y′(0) = 0 .

In the problems 158. to 165. solve the linear differential equations by the method of variation of constants.


158. y′′ − 6y′ + 9y = 9x2+6x+2

x3. 159. y

′′ − 2y′ + y = ex

x.

160. y′′ + y = 1

sinx. 161. y

′′ + 4y = 1sin 2x

.

162. y′′ + y = − cotg2 x . 163. y

′′ + y = 1cos3 x

.

164. y′′ + y = 2

sin3 x. 165. y

′′ + y = tg x .

In the problems 166. to 177. solve the systems by elemination.

166.x = 2x+ y,y = 3x+ 4y . 167.

x = x− 3y,y = 3x+ y .

168.x = −x+ y,y = 3x+ y . 169.

x = 4x− y,y = x+ 2y .

170.x = y + t,y = x− t . 171.

x = −x+ y + et,y = x− y + et .

172.x = −2x− y + sin t,y = 4x+ 2y + cos t . 173.

x = 4x− y − 5t+ 1,y = x+ 2y + t− 1 .

174.x = −5x+ 2y + 40et,y = x− 6y + 9e−t . 175.

x = y − cos t,y = −x+ sin t .

176.x = −5x+ 2y + et,y = x− 6y + e−2t . 177.

x = y,y = x+ et + e−t .

178. Find a curve passing through the point A[2, 0] with the property that the length of the tangent segment between the tangent point and the axisOy is 2 .


179. A mass point is moving along a line such that its kinetic energy in time t is directly proportional to the average velocity in the interval from 0 tot. At the time t=0 the path s=0. Show that the motion is uniform .

180. Determine the time required by a body heated up to 100◦C to be cooled to 25◦C in a room with temperature 20◦C, if it is cooled to 60◦C during10 minutes. According to the Newton law the rate of cooling is proportional to the temperature difference .

181. Find all curves for which the tangent segment between the tangent point and the axis ox is split to halves by the axis oy .

182. A tank is filled by 100l of water solution containing 10kg od salt. Water flows into the tank at a rate 3l per 1 minute, the uniform concentrationis maintained by stirring the solution in tank. Calculate the amount of salt in tank after 1 hour .

183. The body of mass 1kg started to move at time t=0s due to action of a force which was directly proportional to the time and indirectly proportionalto the velocity of the body. At the time t=10s, the velocity was v=0.5ms−1 and the force was F=4×10−5N. Determine the velocity of the body afterone minute .

184. Find the curves, whose tangent distance from the origin is equal to the x coordinate of the tangent point .

185. Find the curves, whose arbitrary tangent segment on the axis oy has length equal to the x coordinate of the tangent point .

186. Find the curves, whose arbitrary tangent intersects with the axis ox at a point equally distant both from the origin and from the tangent point .

187. Find the curves, whose tangent at an arbitrary point P forms together with the axis oy an the segment OP , where O is the origin of the coordinatesystem, a triangle of the area a2 .

188. Find the curves, whose tangent at an arbitrary point P0[x0, y0] makes a segment on the axis oy of the length y20 .

189. Find a curve, which passes through the point O and for which the midpoint between arbitrary tangent point and intersection point of the pertinentnormal to the curve at the same point with the axis ox lies on the parabola y2=ax .

190. How much time needs a boat, moving only by own inertia, to slow down to 1cms−1, if its initial velocity is 1.5ms−1. The resistance of water isproportional to the velocity, such that during the first four seconds the boat is slowed down to 1ms−1 .


191. A mass point X of mass m is attracted by each point S1, S2 while the force is directly proportional to the distance from the pertinent point. Findthe equation of motion for the point X knowing that in the initial instant the point X lies on the segment S1S2 at the distance x0 from its midpointand the velocity is zero .

192. A cylindrical wooden block with cross-sectional area S, height h and mass density ρ floats in water. Its axis is vertical. Find the period of theoscillation, which is invoked by pushing the cylinder to the water and subsequently releasing it. Neglect any resistance of the surroundings .

193. A body of mass m moves straightforwardly under the acting constant force F0 – free fall. Simultaneously, the resistance of the surroundings isconsidered which is proportional to the velocity of the motion with proportionality coefficient β. Find the equation of the motion x=x(t), if x(0)=0 andalso velocity is zero at the time t=0. Find also velocity as a function of time .

Conclusion

The solution of several types of ordinary differential equations was discussed in the chapter. We learned to solve the first order differential equations,where the variable can be separated. We also learned to solve the linear first order differential equation. Besides we demonstrated the solution of othertypes of the first order equations: homogeneous, Bernoulli and exact differential equations. For all these equations, we are able to prescribe an initialcondition and to find the solution satisfying this condition. For the case of linear differential equations we discussed also higher order equations thoughonly with constant coefficients. Now, we also know the methods of unique determination of integration constants by the initial conditions. We alsosolved systems of differential equations of the first order which we are able to transform to a sole equation of a higher order. In the final part of thechapter, several word examples and problems were presented which lead to differential equations discussed in the previous parts. In practice, definitionof the problem by words instead of equations and transformation of the word expression to the differential equation with boundary or initial conditionsis exactly what can be applied in studying other disciplines. Of course, crucial is the ability to solve such equations.


References

[1] Šiagová, J.: Advanced Mathematics, Bratislava, STU, 2011.

[2] Krivá, Z.: Mathematics I, Bratislava, STU, 2007.

[3] Bubeník, F.: Problems to Mathematics for Engineers, Praha, ČVUT, 1999.


[5] Neustupa, J., Kračmar S.: Problems in Mathematics I, Praha, ČVUT, 1998.

[6] Neustupa, J.: Mathematics I, Praha, ČVUT, 2004.

Problem solutions

The answers to the self-assessment questions, if you are unable to formulate, and also a lot of other answers can be found in provided references.

1. y = ce√x 2. arctg y = arcsinx+c 3. ex+y = c|(y+1)(x+1)| 4. 1+ey = c(1+x2) 5. arctg ex = 1

2 sin2 y+c 6.

√1 + x2+

√1 + y2 = c

7. y3 = 3x − 3x2 + c 8. y = tg (cx) 9.√x +√y = c 10. ey(x3 − 1)

13 = c(y + 1)2 11. y = 2 sin

√x + c 12. (x − 5)(y + 4) =

c 13. 10x + 10−y = c 14. x2 + y2 = ln cx2 15. y = 11−cx 16. x2(1 + y2) = c 17. y = ce−

1x2 18. ln | y

y+1| = x2

2+ x + c

19. 2(x3−y3)+3(x2−y2) = −5 20. y = 1+x1−x 21. y = 1, x 6= 0 22. y = 1 23. y = e

√x−2 24. cosx =

√2 cos y 25. 2

√ey2 =

√e(1+ex)

26. y = sinx 27. y = e−√1−x2 28. y = 2 sin2 x− 1

229. y = ce−2x+e−x 30. y = cx3−x2 31. y = c−e−x2

2x232. y = lnx+ c

x33. y =

(x+1)2(12x2 +x+c 34. y = cx+x2 35. y = c−cos 2x

2 cosx36. y = c cosx+sinx 37. y = e2x

4+ce−2x 38. y = x3

4+ c

x39. y = x2 sinx+cx2

40. y = ce− arctg x + arctg x− 1 41. y = c sinx− 2 42. y = (x2 + 1)(c+ x) 43. y = 23

tg x(3− sin2 x) + c3 cosx

44. y = (c− cosx)√x2 + 1

45. y = c+sinxx+1

− cosx 46. y = ln(c lnx − 1) 47. y = e1−3x + x3− 1

948. y = x2 49. y = 2

x2− 1

x350. y = (ex − 1)

√x+ 1

51. y = 2x−sin 2x+4−π4 sinx

52. y = 12(x(x + 1) −

√1+x1−x arcsinx) 53. tg y

x= ln cx 54. y = ce

yx 55. e−

yx = − ln cx 56. y = xe1+cx

57. y − 2x = cx3(y + x) 58. y2 = x(x − c) 59. sin yx

+ ln |x| = c 60. y +√x2 + y2 = cx2 61. y = 2x ln |x| + cx 62. x2e

x2

y2 = c

63. x2 − 2xy + 2y2 = c 64. x2 + y2 = cy 65. sin yx

= cx 66. y2 = 2x2 ln |cx| 67. y = cexy 68. 2y2 ln y − x2 = 0 69. y = x2−1

2

70. y = x2−x 71. eyxarctg y

x =√x2 + y2 72. x

2

y2= 1−2 ln |y| 73. y = 1+

√1 + 2x− x2 74. y = 1

1+cex2, y = 0 75. y2(cex2+1) = 1, y = 0

76. xy(ln2 x + c) = −2, y = 0 77. y = (c√

e−x + x − 2)2, y = 0 78. y = (√ce−x + x− 1 79. y2 lnx = c + sin x, y = 0 80. y =

(c−cosx)2x2

, y = 0 81. y2(c−x) sinx = 1, y = 0 82. y2(cex+x) = 1, y = 0 83. y3 = ce4x x4− 5

1684. y2 = cx2−x 85. y = 1

cx+lnx+1, y = 0


86. y = 24−3ex 87. y =

√2√

1− x2 + x2 − 1 88. y = 1x(1−lnx) 89. y = 1

16cos2 x(2 sinx −

√2)2 90. x3 + y3 − x2 − xy + y2 = c

91. 3y2x + x2 + x2y + 3y = c 92. x− y2

x= c 93. x4 + x2y2 + y4 = c 94. x3y + x2 − y2 = cxy 95. x + ye

xy = c 96. x + arctg x

y= c

97. (x2 + y2)ex = c 98. x3 + 3x2y − y3 = c 99. xey − y2 = c 100. y = c1e5x + c2 101. y = c1e

3x + c2e−3x 102. y = c1e

x + c2e−4x

103. y = c1e−3x + c2xe−3x 104. y = c1e

2x cos 3x + c2e2x sin 3x 105. y = c1 cos 4x + c2 sin 4x 106. y = c1e

−x cos 2x + c2e−x sin 2x

107. y = c1e2x+ c2e

4x 108. y = c1 cos 5x+ c2 sin 5x 109. y = c1e52x+ c2xe

52x 110. y = c1e

−6x+ c2e12x 111. y = c1e

−x cos 23x+ c2e

x sin 23x

112. y = c1 cosx+c2 sinx 113. y = c1+c2ex+c3e

−x 114. y = c1+c2x+c3ex 115. y = c1e

x+c2xex+c3x2ex 116. y = c1e

−x+c2xe−x+c3e2x

117. y = c1+c2e2x+c3e

−2x 118. y = c1e−2x+c2e

x cos(√

3x)+c3ex sin(

√3x) 119. y = (c1 cosx+c2 sinx)ex+(c3 cosx+c4 sinx)e−x 120. y =

c1 + c2e−3x+ 1

4ex 121. y = c1e

4x+ c2xe4x+ 14(x+1)e2x 122. y = c1e

x cos√

2x+ c2ex sin√

2x+ x3

+ 59

123. y = c1 cosx+ c2 sinx+(2x−2)ex

124. y = c1e2x+c2e

−2x+2e3x 125. y = c1e2x+c2e

x−xex 126. y = c1e−2x+c2e

−2x+ 12x2e−2x 127. y = c1 cosx+c2 sinx−(cosx+ 1

2sinx)x

128. y = c1e2x + c2e

5x + 4 129. y = c1e2x + c2e

5x + 2x2 + 1 130. y = c1e2x + c2e

5x + 6e3x 131. y = c1e2x + c2e

5x + (x2 + 3x)e2x

132. y = c1e2x + c2e

5x + (x3 − x)e5x 133. y = c1e2x + c2e

5x + (3 cosx − sinx)e2x 134. y = c1 + c2e4x3 + 2x4 + 6x3 + 3x2 − 8x

135. y = c1 + c2e4x3 + xe

4x3 136. y = c1 + c2e

4x3 + 4 cosx − 3 sinx 137. y = c1e

x3 + c2xe

x3 + e−

x3 138. y = c1e

x3 + c2xe

x3 + 1

2cos x

3

139. y = c1ex3 + c2xe

x3 + 9x2 + 102x + 451 140. y = c1 cos 2x + c2 sin 2x + 1

4x sin 2x 141. y = c1 cos 2x + c2 sin 2x − 1

5cos 3x 142. y =

c1 cos 2x+ c2 sin 2x+ 18e−2x 143. y = e2x(c1 cosx+ c2 sinx) + e2x 144. y = e2x(c1 cosx+ c2 sinx) + 1

8(cosx+ sinx) 145. y = e2x(c1 cosx+

c2 sinx)− 12e2x sin 2x 146. y = c1e

−x+ c2e2x−2x+1+ex 147. y = c1 + c2e

3x−3x2−2x+cosx+3 sin x 148. y = c1ex+ c2xex− sinxex+2

149. y = (c1 cosx+ c2 sinx)e−x + xex + e−x 150. y = (c1 cosx+ c2 sinx)e2x + cosx− sinx+ e2x + 15

151. y = c1e−2x + c2e

−3x + 12e−x + xe−2x

152. y = x2 + e3x 153. y = cosx+x sinx 154. y = 2 sin x2−6 cos x

2155. y = 2ex+ (sin x−2 cosx)e−x−4 156. y = 1

2e−x+ 3

2ex+x2−2

157. y = 12ex − e2x + 1

2e3x 158. y = c1e

3x + c2xe3x + 1x

159. y = (x ln |x| + c1x + c2)ex 160. y = (c1 + ln | sinx|) sinx + (c2 − x) cosx

161. y = c1 cos 2x+ c2 sin 2x+ 14

cos 2x ln | cos 2x|+ 12

sin 2x 162. y = c1 cosx+ c2 sinx+ cosx ln | tg x2|+ 2 163. y = c1 cosx+ c2 sinx− cos 2x

2 cosx

164. y = c1 cosx + c2 sinx + cos 2x2 sinx

165. y = c1 cosx + c2 sinx + 12

cosx ln∣∣ sinx+1sinx−1

∣∣ 166. x = c1et + c2e

5t, y = −c1et + 3c2e5t 167. x =

et(c1 cos 3t+ c2 sin 3t), y = et(c1 sin 3t− c2 cos 3t) 168. x = −c1e−2t+ 13c2e

2t, y = c1e−2t+c2e

2t 169. x = c1e3t+ c2te

3t, y = c1e3t+ c2(t−1)e3t

170. x = c1et−c2e−t+t−1, y = c1e

t+c2e−t−t+1 171. x = c1+c2e

−2t+et, y = c1−c2e−2t+et 172. x = c1+c2t+2 sin t, y = −2c1−c2(2t+1)− 3 sin t− 2 cos t 173. x = c1e

3t + c2te3t + t, y = c1e

3t + c2(t− 1)e3t− t 174. x = c1e−4t + c2e

−7t + 7et + e−t, y = 12c1e−4t− c2e−7t + et + 2e−t

175. x = c1 cos t+ c2 sin t− t cos t, y = −c1 sin t+ c2 cos t+ t sin t 176. x = c1e−4t + c2e

−7t + 710

et + 15e−2t, y = −c1e−4t + c2e

−7t + 140

et + 310

e−2t

177. x = c1et − c2e−t + 1

2et(t− 1)− 1

2e−t(t+ 1), y = c1e

t − c2e−t + 12t(et + e−t) 178.

√4− x2 + 2 ln |2−

√4−x2x| 180. ≈ 40min 181. y2=cx

182. ≈ 3.9kg 183. ≈ 2.7ms−1 184. x2 + y2 = cx 185. y = cx − x ln |x| 186. x2 + y2 = cy 187. a2 + cy2 = xy 188. y = xc+x

189. y2 = 4ax+ 4a2(1− exa ) 190. ms = −ks; s = v = v0e− k

mt; s = m

kv0(1− e−

km t); t = m

kln v0

v≈ 50s 191. mx = k(d

2− x)− k(d

2+ x); x

- deflection of the point X from the mid point of the segment S1S2 in the direction of the initial deflection x0; d is length of the segment S1S2;equation: x + 2k

mx = 0, x(0) = 0, x(0) = 0; solution: x = x0 cos

√2kmt 192. Sρx = −Sρvxg, x is the deflection of the block from the

equilibrium downwards and ρv is the density of water; equation: x + ρvρgx = 0; solution: x = c1 cosωt + c2 sinωt, ω =

√ρvρg; period T = 2π

ω

193. x+ βmx = F0

m, x(0) = 0, x(0) = 0; x = F0

β(t− m

β(1− e−

βm t)); v = F0

β(1− e−

βm t)


Chapter 5. (Double integral)

Aim

Introduce the students to double integral, concentrating methods of calculation and its applications.

Objectives

1. Determine the type of a domain, defining it geometrically.

2. Understand the notion of double integral on a domain and its calculation using partial integration.

3. Simplify the integrand or integration domain by an appropriate substitution.

4. Learn to apply double integrals in calculating various geometrical and physical characteristics of bodies.

Prerequisities

function; graph of a function; curve; derivative; partial derivative; determinate and indeterminate integral

Chapter 5. (Double integral) – 1

Introduction

This chapter discusses calculation techniques for an integral of a function of two variables on a bounded domain. Many geometrical and physicalquantities which can be met solving engineering problems are defined by integrals. Therefore, the calculation techniques for integrals have to be knownto any engineer, this is also the case of multiple integrals as long as practically all bodies are three dimensional in reality. The physical quantities aredefined on such bodies: as a simple example density of an nonhomogeneous body case can be considered. Although, the case of triple integral is notdiscussed directly, the method explained for the double integral, as a reasonable simplification, are directly applicable also for the spacial integral.

First, an efficient way for calculation of double integral includes an efficient way of the integration domain description. Such a description enable toconvert the double integral into two simple integrals calculated subsequently. It will be also shown that under some condition the order of the integrationcan be interchanges to define a better way of integral calculation.

Second, geometric description of the integration domain is crucial. Frequently, it is advantageous to use a substitution to change such a description tosimplify the calculation. Similarly to the simple integral, the substitution can lead to a simpler integral. Unlike the simple integral, however, where theintegration domain is always an interval, here the substitution is often used to modify and simplify the domain.

Finally, several examples for double integral applications will be provided, such as area of a domain, volume of a body, centroid of a planar region etc..

Double integral – Elementary domain and subsequent integration

The description of an integration domain has to be given in a special from, as it has to be able to provide the bounds of the integrals for each variable.

An elementary domain type [x, y], Fig. 5.1(a), is a set D of points in R2, for which there exist continuous functions g1 and g2 in the interval 〈a; b〉 suchthat for any x∈(a; b) holds g1(x)<g2(x) and which is given as: D={(x, y) ∈ R2 : a ≤ x ≤ b ∧ g1(x) ≤ y ≤ g2(x)}.An elementary domain type [y, x], Fig. 5.1(b), is a set D of points in R2, for which there exist continuous functions h1 a h2 in the 〈c; d〉 such that forany y∈(c; d) holds h1(y)<h2(y) and which is given as: D={(x, y) ∈ R2 : h1(y) ≤ x ≤ h2(y) ∧ c ≤ y ≤ d}.A special case of a domain which belongs to both types [x, y] and [y, x] is also a two dimensional interval or a rectangle 〈a; b〉×〈c; d〉. A domain whichis not an elementary domain has to be divided into several parts such that they are already elementary domains.

Let there is a function f continuous in the elementary domain D of the type [x, y] (pertinent bounding functions are denoted g1(x) and g2(x) andpoints a and b). For each x∈〈a;b〉, there is a continuous function hx(y)=f(x, y) defined in the interval y∈〈g1(x);g2(x)〉. Its integral depending on

the parameter x is∫ g2(x)

g1(x)

hx(y) dy, and it is a function F (x) continuous in the interval x∈〈a;b〉. The function F (x) is integrable in the interval 〈a;b〉


O a b

g1(x)

g2(x)y

x

D

O

c

d

h1 (y

)

h2 (y

)

x

y

D

(a) (b)

Figure 5.1: Elementary domains: (a) type [x, y], (b) type [y, x].

renderingb∫

a

F (x) dx =

b∫a

g2(x)∫g1(x)

hx(y) dy

dx =

b∫a

g2(x)∫g1(x)

f(x, y) dy

dx.

The last integral contains subsequent integration. Similarly, subsequent integration of a function f(x, y) on an elementary domain of the type [y, x] canbe introduced.

If the elementary domain D is simultaneously of both types [x, y] and [y, x] and the function f(x, y) is continuous in D, then the order of integrationcan be interchanged, i.e.

b∫a

g2(x)∫g1(x)

f(x, y) dy

dx =

d∫c

h2(y)∫h1(y)

f(x, y) dx

dy.

Example 1 Interchange the order of integration in the integral∫∫D

f(x, y) dx dy=

∫ 2

1

(∫ 2x2

1

f(x, y) dy

)dx considering f a continuous function in

the domain D.

Solution. The subsequent integration is calculated first for the variable y with fixed x (e.g. with x1 along the drawn segment) and only then for thevariable x, the domain D is thus defined as elementary, type [x, y]. It can be written as D = {(x, y) ∈ R2 : 1 ≤ x ≤ 2 ∧ 1 ≤ y ≤ 2x2} and it is draw


in left graph of Fig. 5.2. The domain has to be rewritten as an elementary of the type [y, x], in order to interchange the integration order. The left

O

y=2x2

x2x11

y

8

2y1

y2

1

D

O

x=

√y

2

x21

y

8

21

D1

D2

Figure 5.2: Interchanging the integration order.

graph of Fig. 5.2 confirms that the domain D is not elementary of the type [y, x]. The point of the domain have y coordinates in the interval 〈1; 8〉,nevertheless for y ∈ 〈1; 2〉 the x coordinates of pertinent points lie in the interval 〈1; 2〉 (e.g. for y1), while for y ∈ 〈2; 8〉 x depends on y as

√y2≤ x ≤ 2.

The domainD is to be split int two subdomains D1, D2, Fig. 5.2, right. These domain are written as D1 = {(x, y) ∈ R2 : 1 ≤ x ≤ 2 ∧ 1 ≤ y ≤ 2}and D2 = {(x, y) ∈ R2 :

√y2≤ x ≤ 2 ∧ 2 ≤ y ≤ 8}. They both sum to whole D D = D1 ∪ D2 and do not intersect in the interior. According to the

property V2 (see page Chapter 5. (Double integral) – 5) we obtain

∫∫D

f(x, y) dx dy =

2∫1

2x2∫1

f(x, y) dy

dx =

2∫1

2∫1

f(x, y) dx

dy +

8∫2

2∫√

y2

f(x, y) dx

dy,

and the order of integration is interchanged (i.e. first x, second y).

Double integral – Double integral, the Fubini theorem

The character of physical and geometrical quantities is usually closely related with the domain, with no need to know about the type of the domain.Thus the integral is refferd to as integral over a planar domain.


The double integral over a closed bounded domain D of a function f(x, y) which is bounded on D is a generalization of simple (definite) integral∫ b

a

k(x) dx of a function k(x). The integral is denoted∫∫

Df(x, y) dx dy. We will assume that D is closed bounded planar domain in what follows.

Calculation of the double integral is usually based on he method of subsequent integration using the following Fubini theorem: Let the function f iscontinuous on the domain D, which is an elementary domain of the type [x, y] or [y, x]. Then, it holds respectively: platí

∫∫D

f(x, y) dx dy =

b∫a

g2(x)∫g1(x)

f(x, y) dy

dx

or ∫∫D

f(x, y) dx dy =

d∫c

h2(y)∫h1(y)

f(x, y) dx

dy.

Basic properties additivity and homogeneity of the double integrals are the same as we learned for the simple integrals. Providing that pertinent integralsin the following formulas, the properties are:

V1 :∫∫D

(c1 f1(x, y) + c2 f2(x, y)) dx dy = c1

∫∫D

f1(x, y) dx dy + c2

∫∫D

f2(x, y) dx dy,

V2 :∫∫D

f(x, y) dx dy =

∫∫D1

f(x, y) dx dy +

∫∫D2

f(x, y) dx dy, D = D1 ∪ D2,

where the domains D1 and D2 do not have common inner points. The properties can be extended to a finite number of functions or domains.

Converting the double integral to two subsequently applied simple integrals, the method of definite integrals known from the previous course can beapplied, e. g. the method of substitution, integration by parts etc..

Example 2 Calculate the integral∫∫D

(2x+ 3y2 − 7xy) dx dy, where D = 〈0; 2〉 × 〈−1; 1〉.


Solution. The function f(x, y) = 2x+ 3y2 − 7xy is continuous in the interval D, which is an elementary domain of the type [y, x], thus it is integrablein this domain. The Fubini theorem renders

∫∫D

(2x+ 3y2 − 7xy) dx dy =

1∫−1

2∫0

(2x+ 3y2 − 7xy) dx

dy =

=

1∫−1

[x2 + 3y2x− 7

2x2y

]x=2

x=0

dy =

1∫−1

(4 + 6y2 − 14y) dy =[4y + 2y3 − 7y2

]1−1 = 12.

Remark 1 As long as the domain D is also of the type [x, y], the same theorem provide alternative calculation

∫∫D

(2x+ 3y2 − 7xy) dx dy =

2∫0

1∫−1

(2x+ 3y2 − 7xy) dy

dx =

2∫0

[2xy + y3 − 7

2xy2]y=1

y=−1dx =

2∫0

(4x+ 2) dx =[2x2 + 2x

]20= 12.

The result does not depend on the order of integration.

Example 3 Calculate∫∫D

(x2 + y) dx dy, where D = {(x, y) ∈ R2 : 0 ≤ x ∧ 2x2 ≤ y ≤√x}.

Solution. First, let us plot the domain D. Its boundary is formed by parts of two parabolas with the equations y = 2x2 and y =√x, Fig. 5.3. The

intersection points of the curves are the points O = (0, 0) and P = (3√22,

3√42), obtained by solving the system of equations

y = 2x2, y =√x.

The domain D shown in the picture can be seen as an elementary domain of the type [x, y] defined by the inequalities

0 ≤ x ≤3√2

2, 2x2 ≤ y ≤

√x.


O

P

y=2x2

y=√x

y

3√4

2

x3√2

2

D

Figure 5.3: The domain D = {(x, y) ∈ R2 : x ≥ 0 ∧ 2x2 ≤ y ≤√x}.

As the function f is continuous on D, hence integrable, the Fibini theorem enables the following calculation:

∫∫D

(x2 + y) dx dy =

3√22∫

0

√x∫

2x2

(x2 + y) dy dx =

3√22∫

0

[x2y +

y2

2

]y=√xy=2x2

dx =

3√22∫

0

(x

52 +

x

2− (2x4 + 2x4)

)dx =

27

280 3√2.

Remark 2 The domain D in the last example is also of the type [y, x]. The pertinent inequalities are then

0 ≤ y ≤3√4

2, y2 ≤ x ≤

√y

2,


and according to the Fubini theorem we can alternatively treat it as:

∫∫D

(x2 + y) dx dy =

3√42∫

0

√

y2∫

y2

(x2 + y) dx

dy =

3√42∫

0

[x3

3+ yx

]x=√ y2

x=y2dy =

=

3√42∫

0

(y

32

232 3

+y

32

212

−(y6

3+ y3

))dy =

[7

15√2y

52 − y7

21− y4

4

] 3√42

0

=27

280 3√2.

The result is the same.

Double integral – The method of substitution

Sometimes the integration domain cannot be easily described as an elementary domain or as a union of elementary domains. In such a case, the domainhas to be appropriately transformed.

The method of substitution is used, as well as for the simple definite integral, to simplify the calculation. Here, however, we have two possibilities: first,to simplify the function, second, to simplify the domain.

Let there is an injective (possibly, besides the boundary) and regular mapping g=(gx, gy) determined by the equations x=gx(ξ, η), y=gy(ξ, η) definedon the domain D∗⊂R2. Let the image of the domain D∗ is a domain D, i. e. g : D∗ 7→ D. Then for a continuous function f(x, y) on D holds∫∫

D

f(x, y) dx dy =

∫∫D∗

f(gx(ξ, η), gy(ξ, η)) |Jg(ξ, η)| dξ dη,

where Jg(ξ, η) is determinant

Jg(ξ, η) =

∣∣∣∣∣∣∣∣∂gx(ξ, η)

∂ξ

∂gx(ξ, η)

∂η

∂gy(ξ, η)

∂ξ

∂gy(ξ, η)

∂η

∣∣∣∣∣∣∣∣called jacobian of the mapping g.


The most frequent substitution is the polar transformation, or sometimes generalized polar transformation. The relations required for calculation of theintegrals by this substitution are summarized in Tab. 5.1. The table also provides the types of domains which are transformed to the simplest possibledomain — interval. The pictures give us a guess of domains suitable for an appropriate transformation.

Polar coordinates (ξ, η)↔ (ρ, ϕ) A domain D mapped to an interval

(x, y) ∈ R2, (ρ, ϕ) ∈ 〈0;+∞)× 〈−π; π〉

x = ρ cosϕ, y = ρ sinϕ

gx(ρ, ϕ) = ρ cosϕ, gy(ρ, ϕ) = ρ sinϕ

Jg(ρ, ϕ) = ρ

(identifying the polar axis p with the axis x)

∫∫D

f(x, y) dx dy =

∫∫D∗

f(ρ cosϕ, ρ sinϕ) ρ dρ dϕ

O x≡p

D

ρ=const.

ρ=const.y

ϕ=

cons

t.

ϕ=con

st.

Generalized polar coordinates (ξ, η)↔ (ρ, ϕ) A domain D mapped to an interval

(x, y) ∈ R2, (ρ, ϕ) ∈ 〈0;+∞)× 〈−π; π〉

x = aρ cos ϕ, y = bρ sin ϕ, a > 0, b > 0

gx(ρ, ϕ) = aρ cos ϕ, gy(ρ, ϕ) = bρ sin ϕ

Jg(ρ, ϕ) = abρ

(identifying the polar axis p with the axis x)

∫∫D

f(x, y) dx dy =

∫∫D∗

f(aρ cos ϕ, bρ sin ϕ) abρ dρ dϕO x≡p

D

ρ=const.

ρ=const.

y

ϕ=

cons

t.

ϕ=con

st.

Table 5.1: Substitutions for the double integrals.


Example 4 Calculate∫∫D

e−x2−y2 dx dy, where D = {(x, y) ∈ R2 : 4 ≤ x2 + y2 ≤ 9 ∧ |y| ≤ x}.

Solution. First, we draw the domain D such that we draw the boundary curves. They are:

circles: x2 + y2 = 4, x2 + y2 = 9, lines: x = y, x = −y,

see Fig. 5.4(left). In this case, the polar transformation x = ρ cosϕ, y = ρ sinϕ is suitable, because the D is the image of an interval. We can verify

O O

g(ρ, ϕ)

x ρ32 32

y ϕ3

2

ρ=3ρ=2

ϕ=π4

ϕ=−π4−3

−2

π4

−π4

D

D∗

Figure 5.4: The domain D = {(x, y) ∈ R2 : 4 ≤ x2 + y2 ≤ 9 ∧ |y| ≤ x} and the domain D∗ = 〈2; 3〉 × 〈−π4; π4〉, g : D∗ 7→ D.

this by substituting the transformation equations int the equations of the boundary curves. We obtain

x2 + y2 = ρ2 cos2 ϕ+ ρ2 sin2 ϕ = ρ2,

x = ±y ⇒ ρ cosϕ = ±ρ sinϕ⇒ cosϕ = ± sinϕ⇒ tgϕ = ±1.

It implies that the boundary curves can be expressed in polar coordinates as ρ = 2, ρ = 3, ϕ = π4, ϕ = −π

4. The domain D∗ bounded by these curves

is an interval, i. e. D∗ = {(ρ, ϕ)∈R2 : ρ ∈ 〈2; 3〉 ∧ ϕ ∈ 〈−π4; π4〉}. The mapping g(ρ, ϕ) = (ρ cosϕ, ρ sinϕ) maps the interval D∗ onto the domain

D. This mapping is also shown in Fig. 5.4, where the domain D∗ is drawn in a rectangle coordinate system Oρϕ such that x is replaced by ρ and y isreplaced by ϕ. In this way the domain D∗ is also optically represented by an interval in the coordinate system Oρϕ.


As the function f(x, y) = e−x2−y2 is continuous on the domain D, the integral exists. Using the method of substitution, we obtain

∫∫D

e−x2−y2 dx dy =

∫∫D∗

e−ρ2 cos2 ϕ−ρ2 sin2 ϕ ρ dρ dϕ =

π4∫

−π4

3∫2

e−ρ2

ρ dρ dϕ =

π4∫

−π4

dϕ

3∫2

e−ρ2

ρ dρρ2=t=

π

2

9∫4

e−t1

2dt =

π

4

(e−4 − e−9

).

Remark 3 If the direct integration were used, the domain would be necessarily split int several parts in order to have been able to find theintegration bounds and to use the Fubini theorem.

Example 5 Caculate by the method of substitution∫∫D

xy2 dx dy, where

D = {(x, y) ∈ R2 : x ≥ 0 ∧ y ≥ 0 ∧ 1 ≤ xy2 ≤ 4 ∧ x ≤ y3 ≤ 8x}.

Solution. Though, in the most of the problems we solve, the polar coordinates are used, there exist examples where other substitution is advantageous.This is one of them.

The integration domain is drawn in Fig. 5.5(left). The domain look trivial even for direct integration and we would be able to find the integrationbounds for the subsequent integration in the Fubini theorem. Nevertheless, an appropriate transformation can simplify the calculation. Let us considerthe curves with the equations

xy2 = 1, xy2 = 4,y3

x= 1,

y3

x= 8,

which form the contour of the domain D. In this case a power transformation can be effectively used:

(x, y) ∈ (0;+∞)× (0;+∞), (u, v) ∈ (0;+∞)× (0;+∞) : x = uα1 vβ1 , y = uα2 vβ2 ,

where α1, α2, β1, β2 ∈ R, and α1β2 − α2β1 = κ 6= 0,

with jacobian Jg(u, v) = κuα1+α2−1 vβ1+β2−1 and inverse mapping u = xβ2κ y−

β1κ , v = x−

α2κ y

α1κ .

We define new variables u and v as

u = xy2, v =y3

x,


O O

g(u, v)

xy2=1

xy2=4

y3

x=1

y3

x=8

x u1 4

y v

8

1

D T

Figure 5.5: The domain D = {(x, y) ∈ R2 : x ≥ 0 ∧ y ≥ 0 ∧ 1 ≤ xy2 ≤ 4 ∧ x ≤ y3 ≤ 8x} and the domain T = 〈1; 4〉 × 〈1; 8〉, g : T 7→ D.

because either u or v are then constant on the boundary. The coefficients of the power transformation α1, α2, β1, β2 can be determined by the conditions

β2κ

= 1, −β1κ

= 2, −α2

κ= −1, α1

κ= 3,

which givesβ2 = κ, β1 = −2κ, α2 = κ, α1 = 3κ,

κ = α1β2 − α2β1 = 3κ2 + 2κ2 = 5κ2 ⇒ κ =1

5,

β2 =1

5, β1 = −

2

5, α2 =

1

5, α1 =

3

5.

Our power transformation is in the present case defined by the relations

x = u35v−

25 , y = u

15v

15 , Jg(u, v) =

1

5u−

15v−

65

and it maps the elementary domain T = 〈1; 4〉× 〈1; 8〉 onto the domain D, see Fig. 5.5 — the figure also shows that replacing x by u and replacing y byv, presents T as an interval in the coordinate system Ouv. The function f(x, y) = xy2 is continuous on D which implies its integrability. The methodof substitution renders∫∫

D

xy2 dx dy =

∫∫T

u |Jg(u, v)| du dv =1

5

8∫1

4∫1

u45 v−

65 du dv =

1

5

[u

95

95

]41

·

[v−

15

−15

]81

= · · · =5√8

9· 8

5√8− 1

5√8− 1

.


Double integral – Double integral applications

Integral in general is used to calculate geometrical and physical quantities. There is a lot of such examples, only the most principal will be discussed.

The double integral can be used for various quantities which characterize a planar domain D, e.g.:

Area of a domain D: A =

∫∫D

dx dy.

Area of a curved surface P: A =

∫∫D

√1 +

(∂f

∂x

)2

+

(∂f

∂y

)2

dx dy, where D is an orthogonal projection of the surface P, given as a graph of a

finction z = f(x, y), to the plane Oxy.

Volume of a cylindrical body T: V =

∫∫D

f(x, y) dx dy, where D is an orthogonal projection of the body T to the plane Oxy. The body T occupies

the space between the domain D and the graph of the function f(x, y) which is defined on D and satisfying condition f(x, y) ≥ 0 on D.

Mass of a domain D: m =

∫∫D

σ dx dy, where σ(x, y) is the (surface) density of the body defined by the domain D.

Static moments of a domain D with respect to the coordinate axes x, y:

Ux =

∫∫D

y σ(x, y) dx dy, Uy =

∫∫D

x σ(x, y) dx dy.

The centroid T of a domain D: T =

(Uym,Uxm

).


Moments of inertia of a domain D with respect to the coordinate axes x, y and the polar moment:

Jx =

∫∫D

y2 σ(x, y) dx dy, Jy =

∫∫D

x2 σ(x, y) dx dy, Jr =

∫∫D

(x2 + y2)σ(x, y) dx dy.

Example 6 Calculate area of a domain D which is bounded by a curve given by the equation (x2 + y2)3 = a2(x4 + y4) with a parameter a > 0.

Solution. The planar body, whose area is to be calculated, is drawn in Fig. 5.6(left). To calculate the integral∫∫

Ddx dy, the domain is transformed by

O

A

ωx

y

D O ρ

ϕ

ω A

π

−π

D∗ ρ=r(ϕ

)

g(ρ, ϕ)

Figure 5.6: The domain D boundaed by the curve (x2 + y2)3 = a2(x4 + y4) and the domain D∗, g : D∗ 7→ D.

polar coordinates. To this end, we substitute the relations x = ρ cosϕ, y = ρ sinϕ to the equation of the boundary curve for expressing it in a suitableform. We obtain

(ρ2 cos2 ϕ+ ρ2 sin2 ϕ)3 = a2(ρ4 cos4 ϕ+ ρ4 sin4 ϕ)

which makesρ2 = a2(cos4 ϕ+ sin4 ϕ)⇒ ρ = a

√cos4 ϕ+ sin4 ϕ,

since both ρ and a are nonnegative.


The polar coordinates has an explicit geometric meaning. The last relation for example gives the distance ρ of a point on the curve from the beginningof the coordinate system depending on the angle ϕ. Hence, if we choose ϕ = ω ∈ 〈−π; π〉 (the halfline

−→OA), see Fig. 5.6(left), the points of the domain

D lying on the segment OA satisfy the relationρ ≤ a

√cos4 ω + sin4 ω.

We denote g the mapping which maps the domain D∗

D∗ ={(ρ, ϕ) : ϕ ∈ 〈−π; π〉 ∧ ρ ∈ 〈0; r(ϕ)〉, r(ϕ) = a

√cos4 ϕ+ sin4 ϕ

}to the domain D. Then Fig. 5.6 presents the domain D∗ in a rectangular coordinate system Oρϕ, where it is an elementary domain of the type [ϕ, ρ]mapped by g onto D.Using the Fubini theorem, the double integral can be calculated as follows:

A =

∫∫D

dx dy =

∫∫D∗

ρ dρ dϕ =

π∫−π

r(ϕ)∫0

ρ dρ

dϕ =

π∫−π

1

2

(a

√cos4 ϕ+ sin4 ϕ

)2

dϕ =a2

2

π∫−π

(cos4 ϕ+ sin4 ϕ

)dϕ =

=a2

2

π∫−π

(1 + cos 2ϕ

2

)2

+

(1− cos 2ϕ

2

)2

dϕ =a2

2

π∫−π

(1

4+

1

2cos 2ϕ+

1

4cos2 2ϕ

)+

(1

4− 1

2cos 2ϕ+

1

4cos2 2ϕ

)dϕ =

=a2

2

π∫−π

(1

2+

1

2cos2 2ϕ

)dϕ =

a2

2

π∫−π

(1

2+

1

2

(1 + cos 4ϕ)

2

)dϕ =

a2

2

π∫−π

(3

4+

1

4cos 4ϕ

)dϕ =

3

4πa2,

because∫ π

−πcos 4ϕ dϕ = 0 due to appropriately oscillating character of the periodic function.

Example 7 Calculate area of a part of a conical surface P, with P given as

P = {(x, y, z) ∈ R3 : z =√x2 + y2 ∧ z ≤

√2(x2+ 1)}.


x

y

z

PO x

y

2 2+2√22−2

√2

2

−2

D

Figure 5.7: The orthogonal projection of the surface P to the plane Oxy is the domain D.

Solution. A part of the conical surface P with the equation z = f(x, y) =√x2 + y2 is drawn in Fig. 5.7(left) together with a part of the plane

z =√2(x2+ 1). The curved surface, whose area is to be calculated, lies below the given plane as shown in the picture. The orthogonal projection of

the surface to the plane Oxy has to be found first. Let us denote this projected domain D. We substituted z from the equation of the plane to theequation of the cone. It renders √

2(x2+ 1)=√x2 + y2,

which can be squared to come to

2

(x2

4+ x+ 1

)= x2 + y2.

After small arrangements we obtain

x2 − x2

2− 2x− 2 + y2 = 0⇒ x2 − 4x− 4 + 2y2 = 0,

(x− 2)2 + 2y2 = 8⇒(x− 2

2√2

)2

+(y2

)2= 1.

The last equation is an equation of a ellipse with the centre at the point (2, 0) and halfaxes of the lengths 2√2 and 2. The domain D lies in its interior.

This domain is drawn in Fig. 5.7(right). Calculating the partial derivatives of the function f(x, y)

∂f(x, y)

∂x=

x√x2 + y2

,∂f(x, y)

∂y=

y√x2 + y2

⇒

√1 +

(∂f(x, y)

∂x

)2

+

(∂f(x, y)

∂y

)2

=√2,


we come to a simple expression for the area A =

∫∫D

√2 dx dy. The domain D is an ellipse and such a case is appropriate for using generalized polar

coordinates. This coordinates have to be shifted such that the origin is at the point (2, 0). The substitution relations for the composed transformationwith parameters a and b from the polar transformation given by a = 2

√2 and b = 2, respectively, read

x = gx(ρ, ϕ) = 2 + 2√2ρ cos ϕ, y = gx(ρ, ϕ) = 2ρ sin ϕ, Jg = 4

√2ρ.

We substitute them to the equation of the ellipse with the result(2 + 2

√2ρ cos ϕ− 2

2√2

)2

+

(2ρ sin ϕ

2

)2

= 1⇒ ρ = 1,

so that the domain D∗ which is mapped to the domain D is an interval D∗ = {(ρ, ϕ) ∈ R2 : ρ ∈ (0; 1〉 ∧ ϕ ∈ 〈−π; π)}. Now, the substitution methodcan be used and the area is calculated

A =

∫∫D

√2 dx dy =

∫∫D∗

√2 4√2 ρ dρ dϕ = 8

1∫0

π∫−π

ρ dϕ dρ = 81

22π = 8π.

Remark 4 The area could be calculated directly in rectangular coordinates which would save us from the substitution. Nevertheless, theintegration would be more complicated. We can compare both methods:

A =

∫∫M

√2 dx dy = 2

2+2√2∫

2−2√2

√2+2x−x2

2∫0

√2 dy dx = 2

2+2√2∫

2−2√2

√4 + 4x− x2 dx =

= 2

2+2√2∫

2−2√2

√8− (x− 2)2 dx = 2

√8

2+2√2∫

2−2√2

√1−

(x− 2

2√2

)2

dxt=x−2

2√2

= 2√8

1∫−1

√1− t2

√8 d t =

= 8[t√1− t2 + arcsin t

]1−1

= 8(arcsin 1− arcsin(−1)) = 8π.


Example 8 Calculate volume of a body T bounded by the paraboloid of rotation z = x2 + y2 an the planes z = 0, y = 1, y = 2x and x+ y = 6.

Solution. The given body T is bounded from below by the plane Oxy and from above by the given paraboloid. The rest of the boundary is formed bythree vertical planes, so that the orthogonal projection of the body T to the plane Oxy is the domain D drawn in Fig. 5.8. The volume V of the body

Ox6521

2

y

6

4

1D y=1

y=2x

y=6−x

Figure 5.8: The domain T for calculation of the body T volume.

T is in such a case given by simple integration of the upper bound

V =

∫∫D

(x2 + y2) dx dy =

4∫1

6−y∫y2

(x2 + y2) dx dy =

4∫1

[x3

3+ y2x

]6−yy2

dy =

4∫1

(−15

8y3 + 12y2 − 36y + 72

)dy =

2511

32.

Example 9 Find the coordinates of the centroid of a thin homogeneous plate, whose form is a circular sector of an angle α and a radius R.

Solution. The domain D of the thin plate is shown in Fig. 5.9. The plate is homogeneous so its density σ is a constant. We need to calculate the massm of the domain D and its static moments Sx and Sy with respect to coordinate axes. Let us choose the coordinate system Oxy as it is shown in thepicture. We have a free choice for this coordinate system so it is worth thinking about the most advantageous position utilizing the shape of the domain.The mass of the domain is obtained by the substitution with the polar coordinates

m =

∫∫D

σ dx dy =

α2∫

−α2

R∫0

σ ρ dρ dϕ = σ

α2∫

−α2

dϕ

R∫0

ρ dρ = αR2

2σ.


OxR

yR sin α

2

−R sin α2

Dα2

α2

Figure 5.9: The domain D for finding the centroid.

The static moment with respect to the y axis is

Sy =

∫∫D

σ x dx dy =

α2∫

−α2

R∫0

σ ρ2 cosϕ dρ dϕ = σ

α2∫

−α2

cosϕ dϕ

R∫0

ρ2 dρ = sinα

2

2R3

3σ.

The homogeneous domain D is symmetric with respect to the axes x so the pertinent static moment should be zero. We can verify it by calculating theintegral

Sx =

∫∫D

σ y dx dy =

α2∫

−α2

R∫0

σ ρ2 sinϕ dρ dϕ = σ

α2∫

−α2

sinϕ dϕ

R∫0

ρ2 dρ = 0.

Hence, the coordinates of the thin plate T centroid are

Tx =Sym

=

(sin α

2

)2R3

3σ

αR2

2σ

=4

3Rsin α

2

α, Ty =

Sxm

= 0.

It means that the centroid lies on the axes of symmetry of the plate and its distance from the vertex of the sector is given by the value of Tx


Self-assessment questions and problems for solving

1. Describe common geometrical shapes as elementary domains, e.g. a circle, a circular segment, an ellipse, a cardioid etc.. How do you proceed in thedescription?

2. Cut a rectangular hole in aforementioned shapes and split the resulting domain to elementary domains. How do you proceed in this case?

3. Could you integrate a discontinuous function on a given domain?

4. Can you imagine how a three dimensional integral (a triple integral) is calculated based on your experience with the double integral?

5. Are you able to integrate on a map with the system of contour and descent lines by the method of substitution?

6. Resuming your knowledge from other fields, explain which physical quantities have “integral” character and thus are calculated by the double integral?

In the problems 1. to 10. calculate subsequently the integrals.

1.

1∫0

1∫0

x2

1 + y2dx

dy . 2.

2∫0

1∫0

(x2 + xy + x− y2) dx

dy .

3.

2∫1

1∫0

(1 + xy + x2 + y2) dx

dy . 4.

1∫0

dx

x∫x2

xy2 dy .

5.

π∫0

dx

1+cosx∫0

y2 sinx dy . 6.

2∫1

x2∫1x

(x+ y) dy

dx .

7.

π2∫

0

2 cosϕ∫0

(1 + ρ) dρ

dϕ . 8.

π∫0

dϕ

3 sinϕ∫0

ρ cosϕ dρ .


9.

1∫12

1∫x

v + 1

vdv

du . 10.

π2∫

0

1∫cos y

x4 dx

dy .

In the problems 11. to 18. change the order of integration.

11.

2∫1

4∫3

f(x, y) dx

dy . 12.

π∫0

sinx∫0

f(x, y) dy

dx .

13.

2∫0

6−x∫2x

f(x, y) dy

dx . 14.

1∫−1

√1−x2∫0

f(x, y) dy

dx .

15.

1∫0

√x∫

x3

f(x, y) dy

dx . 16.

1∫0

√

1−y2∫−√

1−y2

f(x, y) dx

dy .

17.

2∫−6

2−y∫14y2−1

f(x, y) dx

dy . 18.

2∫0

√2x∫

√2x−x2

f(x, y) dy

dx .

In the problems 19. to 32. calculate the double integrals.

19.∫∫D

1

x+ ydx dy, where D = 〈0; 1〉 × 〈1; 2〉 .

20.∫∫D

x2 + y2 − 2x− 2y + 4dx dy, where D = 〈0; 2〉 × 〈0; 2〉 .


21.∫∫D

ln(1 + x)2y dx dy, where D = 〈0; 1〉 × 〈0; 1〉 .

22.∫∫D

x2y sin(xy2) dx dy, where D = 〈1; 2〉 × 〈0; π2〉 .

23.∫∫D

x2 − y2

(x2 + y2)2dx dy, where D = 〈0; 1〉 × 〈0; 1〉 .

24.∫∫D

(2x+ y) dx dy, where D = {(x, y) ∈ R2 : x ≥ 0 ∧ y ≥ 0 ∧ x+ y ≤ 3} .

25.∫∫D

xy dx dy, where D = {(x, y) ∈ R2 : x ≥ 0 ∧ y ≥ 0 ∧√x+√y ≤ 1} .

26.∫∫D

|x| dx dy, where D = {(x, y) ∈ R2 : x2 ≥ y ∧ 4x2 + y2 ≤ 12} .

27.∫∫D

2x

x2 + y2dx dy, where D = {(x, y) ∈ R2 : x ≥ 0 ∧ y ≤ 1 ∧

√x ≤ y} .

28.∫∫D

x2 y2 dx dy, where D = {(x, y) ∈ R2 : 1 ≤ x2 + y2 ≤ 4} .

29.∫∫D

arctgx

ydx dy, where D = {(x, y) ∈ R2 : 1 ≤ x2 + y2 ≤ 9 ∧ x√

3≤ y ≤ x

√3} .


30.∫∫D

(x+ y + xy) dx dy, where D is a domain bounded by the curve y = 4− x2 and by the x axis .

31.∫∫D

x2

y2dx dy, where D is a domain bounded by the curve xy = 1 and by the lines x = 2 and y = x .

32.∫∫D

exy dx dy, where D is a domain bounded by the parabola y2 = x and by three lines, one of the being the y axis and the other two are parallel

to the x axis passing respectively through the points (0, 1) and (0, 2) .

In the problems 33. to 40. calculate the integrals by the substitution with polar coordinates.

33.∫∫D

(1− 2x− 3y) dx dy, where D = {(x, y) ∈ R2 : x2 + y2 ≤ 4 ∧ y ≤ x} .

34.∫∫D

√x2 + y2 dx dy, where D = {(x, y) ∈ R2 : 0 ≤ y ≤ x ∧ 0 ≤ x ≤ 1} .

35.∫∫D

ln(x2 + y2)

x2 + y2dx dy, where D = {(x, y) ∈ R2 : 1 ≤ x2 + y2 ≤ e ∧ 0 ≤ y} .

36.∫∫D

sin√x2 + y2 dx dy, where D = {(x, y) ∈ R2 : π2 ≤ x2 + y2 ≤ 4π2} .

37.∫∫D

√1− x2 − y2

1 + x2 + y2dx dy, where D = {(x, y) ∈ R2 : x2 + y2 ≤ 1 ∧ 0 ≤ x} .


38.∫∫D

(x+ y) dx dy, where D = {(x, y) ∈ R2 : x2 + y2 ≤ p(x+ y)}, (p > 0) .

39.∫∫D

(xy)2 dx dy, where D = {(x, y) ∈ R2 : (x2 + y2)2 ≤ p2(x2 − y2)} .

40.∫∫D

e−x2−y2 cos(x2 + y2) dx dy, where D = {(x, y) ∈ R2 : x2 + y2 ≤ π

4} .

In the problems 41. to 44. calculate the integrals by the substitution with generalized polar coordinates.

41.∫∫D

(x− 2y) dx dy, where D = {(x, y) ∈ R2 :x2

4+ y2 ≤ 1 ∧ 0 ≤ x ≤

√12y} .

42.∫∫D

√1− x2 − 4y2 dx dy, where D = {(x, y) ∈ R2 : x2 + 4y2 ≤ 1} .

43.∫∫D

ln(1 + x2 + 9y2) dx dy, where D = {(x, y) ∈ R2 : x2 + 9y2 ≤ 36} .

44.∫∫D

(x2 + y2) dx dy, where D = {(x, y) ∈ R2 : 3x2 + 12y2 ≤ 1} .

In the problems 45. to 48. calculate the integrals by an appropriate substitution.

45.∫∫D

xy dx dy, where D = {(x, y) ∈ R2 : x3 ≤ y ≤ 2x3 ∧ 3x ≤ y2 ≤ 4x} .


46.∫∫D

(2x− y) dx dy, where D = {(x, y) ∈ R2 : 1 ≤ x+ y ≤ 2 ∧ 1 ≤ 2x− y ≤ 3} .

47.∫∫D

x2

ydx dy, where D = {(x, y) ∈ R2 : 2x2 ≤ 6y ≤ 3x2 ∧ x ≤ y2 ≤ 7x} .

48.∫∫D

x

ydx dy, where D = {(x, y) ∈ R2 : x ≤ y ≤ 2x ∧ x2 ≤ y ≤ 2x2} .

In the problems 49. to 57. calculate the area of a planar domain bounded by given curves.

49. 2x− y = 0, 2x− y = 7, x− 4y + 7 = 0, x− 4y + 14 = 0 .

50.x2

a2+y2

b2= 1,

x

a+y

b= 1 . 51. y

2 = x, y = x+ 2, y = 2, y = −2 .

52. y =1

a(x− a)2, x2 + y2 = a2, a > 0 . 53. x =

y2 + b2

2b, x =

y2 + a2

2a, a > b > 0 .

54. xy = 6, 3x− y = 7, x = 1 . 55.√x+√y =√a, x+ y = a, a > 0 .

56. 6x = y3 − 16y, 24x = y3 − 16y . 57. xy = 2, xy = 8, y =x

2, y = 2x .

In the problems 58. to 63. calculate the area of a planar domain bounded by given curves using polar coordinates.

58. (x2 + y2)

2= 4(x2 − y2) . 59. (x

2 + y2)2= 2a2xy, a > 0 .

60. (x2 + y2)

3= 2ax5, a > 0 . 61. (x

2 + y2)3= 2axy4, a > 0 .

62. (x+ y)3 = xy . 63. (x2 + y2)

2= a(x3 − 3xy2), a > 0 .


In the problems 64. to 67. calculate the area of a planar domain bounded by given curves using generalized polar coordinates.

64.(x2

a2+y2

b2

)2

=xy

c2, a > 0, b > 0, c > 0 . 65.

(x2

a2+y2

b2

)2

= x2 + y2, a > 0, b > 0 .

66.x2

a2+y2

b2= x+ y, a > 0, b > 0 . 67.

(x2

a2+y2

b2

)2

=x2y

c3, a > 0, b > 0, c > 0 .

In the problems 68. to 75. calculate the area of a curved interface given by the domain D.68. D = {(x, y, z) ∈ R3 : 2x+ 2y + z = 4 ∧ x2 + y2 ≤ 9} .

69. D = {(x, y, z) ∈ R3 : 2z = x2 + y2 ∧ x2 + y2 ≤ 1} .

70. D = {(x, y, z) ∈ R3 : x2 + y2 + z2 = 27 ∧ x2 + y2 ≤ 6z} .

71. D = {(x, y, z) ∈ R3 : x2 + y2 = z2 ∧ z ≥ 0 ∧ z ≤ y√2+√2} .

72. D = {(x, y, z) ∈ R3 : z2 = 2xy ∧ 0 ≤ x ≤ 2 ∧ 0 ≤ y ≤ 4} .

73. D = {(x, y, z) ∈ R3 : y2 + z2 = 4 ∧ x2 + y2 ≤ 4} .

74. D = {(x, y, z) ∈ R3 : z2 = x2 + y2 ∧ z ≥ 0 ∧ x2 + y2 ≤ 2y} .

75. D = {(x, y, z) ∈ R3 : 2z = x2 + y2 ∧ x2 + y2 + z2 ≤ 3} .

In the problems 76. to 85. calculate the volume of a body by the double integral.

76. 2y2 = x, x+ 2y + z = 4, z = 0 . 77. 2y = x2, x+ y + z = 4, z = 0, y = 1 .

78. x2 + y2 = z, x2 + y2 = 2x, z = 0 . 79. x

2 + y2 = 1, y2 + z2 = 1 .

80. z = e−x2−y2 , x2 + y2 = 1, z = 0 . 81. z

2 = 2xy, (x2 + y2)2= 2xy, z = 0 .


82. 2z = 2x2 + 3y2, x2 + 4y2 = 1, z = 0 . 83. (x2 + 4y2)

2+ 9z2 = 1 .

84. z = x2 + y2, y = 2x, y = 6− x, z = 0, y = 1 .

85. z = x2 + y2, x2 + y2 = x, x2 + y2 = 2x, z = 0 .

86. Determine the mass of a planar domain bounded by an ellipse with the equation x2

4+y2

9=1, if the density is in each its point directly proportional

to the distance r from the small halfaxis and for r=1 it is equal to two .

87. Determine the mass of a domain bounded by the parabolas y=2−x2 and y=x2. The density at each point is directly proportional to the distanceof the point from the x axis and it is equal to four at the point (0, 2) .

88. Calculate the mass of a thin plate whose boundary is a curve given by the equation |x|+|y|=π2, if the density σ of the plate is given by the relation

σ(x, y)= cosx cos y .

89. Calculate the mass of a thin plate bounded by the circle x2+y2=2ax. The density σ of the circle is given by the relation σ(x, y)=√x2+y2 .

90. Calculate the static moments of a thin plate bounded by the curves y=|x| and y=√2−x2, if the density in each point is equal to four times the

distance of the point from the x axis .

91. Determine the coordinates of the centroid of a homogeneous domain bounded by the curves y=x2 and y= 21+x2

.

92. Find the centroid coordinates of a homogeneous domain bounded by the curve y=cosx, x∈〈−π; π〉 and by a line parallel to the x axis whichpasses through the point (0, 1) .

93. Calculate the coordinates of the centroid of a planar domain bounded by the curve y=4−x2 and by the x axis. The density at each point isdirectly proportional to the distance from the y axis .

94. Determine the centroid coordinates of a quartercircle given by the relation x2+y2≤2 and lying in the first quadrant, if the density σ is given bythe relation σ(x, y)=2

√2x2+2y2 .

95. The density of a equilateral rectangular triangle with the hypotenuse 2a is directly proportional to the distance from the hypotenuse. Determine


the height of the centroid above the hypotenuse of the triangle .

96. Calculate the coordinates of the centroid of a planar domain bounded by the curve√x+√y=√2 and by the coordinate axes .

97. Calculate the coordinates of the centroid of a planar domain bounded by the parabola y=x2 and by the line y=x+2, if the density in each pointis directly proportional to the distance of the point from the x axis .

98. Determine the moment of inertia of a homogeneous circular domain with the unit density and the radius r=4 with respect to its tangent .

99. A ring domain is bounded by concentric circles with radii r1=1 and r2=2. The density at each point of the domain is directly proportional to thedistance from the centre of the ring. Determine the moment of inertia of the domain with respect to a line passing through the centre of the ring .

100. The density at each point of an annular domain given by the inequalities 1≤x2+y2≤4 is indirectly proportional to the distance of the point fromthe centre of the annular domain. Determine the moment of the inertia of the domain with respect to the x axis .

101. Determine the polar moment of inertia of an ellipse given by the inequality x2+4y2≤4, if the density at each point is directly proportional to itsdistance from the small halfaxis .

102. Calculate the polar inertial moment of the part of the circle x2+y2≤2x+2y which lies in the first quadrant. Consider the circle to be homogeneouswith the unit density .

Conclusion

The chapter was intended to calculation of double integrals. First, we learned to describe the domain of integration such that a double integral isconverted to two subsequently integrated integrals. These simple definite integrals are then calculated by the methods which we learned in the previouscourse of mathematics. We also saw that it is frequently useful to transform the domain of integration by a appropriate function so that the evaluation ofthe integrals were simplified by the substitution method. The final part of the chapter was intended to application of this type of integral in calculation ofvarious geometrical and physical quantities. The acquired knowledge can be applied for understanding and studying other disciplines of our specialization.


References

[1] Šiagová, J.: Advanced Mathematics, Bratislava, STU, 2011.

[2] Krivá, Z.: Mathematics I, Bratislava, STU, 2007.

[3] Bubeník, F.: Problems to Mathematics for Engineers, Praha, ČVUT, 1999.


[5] Neustupa, J., Kračmar S.: Problems in Mathematics I, Praha, ČVUT, 1998.

[6] Neustupa, J.: Mathematics I, Praha, ČVUT, 2004.

Problem solutions

The answers to the self-assessment questions, if you are unable to formulate, and also a lot of other answers can be found in provided references.

1. π12

2. 0 3. 5312

4. 140

5. 43

6. 5610

7. 4+π2

8. 0 9. 138− ln 2 10. 15π−16

15011.

∫ 4

3

∫ 2

1

f(x, y) dy

dx

12.∫ 1

0

∫ π−arcsin y

arcsin y

f(x, y) dx

dy 13.∫ 4

0

∫ y2

0

f(x, y) dx

dy +

∫ 6

4

∫ 6−y

0

f(x, y) dx

dy 14.∫ 1

0

∫ √1−y2

−√

1−y2f(x, y) dx

dy

15.∫ 1

0

∫ 3√y

y2f(x, y) dx

dy 16.∫ 1

−1

∫ √1−x20

f(x, y) dy

dx 17.∫ 0

−1

∫ 2√x+1

−2√x+1

f(x, y) dy

dx+

∫ 8

0

∫ 2−x

−2√x+1

f(x, y) dy

dx

18.∫ 1

0

∫ 1−√

1−y2

12y2

f(x, y) dx

dy+

∫ 1

0

∫ 2

1+√

1−y2f(x, y) dx

dy+

∫ 2

1

∫ 2

12y2f(x, y) dx

dy 19. ln 2716

20. 323

21. 2 ln 2−1 22. 32

23. neexistuje 24. 272

25. 120

26. 4√3− 10

327. ln 2− 2 + π

228. 21π

829. π2

630. 256

1531. 9

432. 1

233. 2π + 8

3

√2

34. 16(√2+ ln(1+

√2)) 35. π

436. −6π2 37. π

4(π−2) 38. 1

2πp3 39. 1

90p6 40. π

241. −2

3(√3−1) 42. π

343. π

3(37 ln 37−36)

44. 5π288

45. 5(2−65 − 1)(4

85 − 3

85 ) 46. 4

347. 5 48. 9

1649. 7 50. ab

2(π2− 1) 51. 40

352. a2

12(3π − 4) 53. 2

3(a − b)

√ab

54. 6 ln 3+2 55. 13a2 56. 16 57. 6 ln 2 58. 4 59. a2 60. 63

128πa2 61. 7

128πa2 62. 1

6063. 1

4πa2 64. a

2b2

2c265. π

2ab(a2+ b2)


66. π4ab(a2 + b2) 67. π

32a5b3

c668. 27π 69. 2

3π(2√2− 1) 70. 18π(3−

√3) 71. 8π 72. 16 73. 32 74.

√2π 75. 2

3π(3√3− 1)

76. 815

77. 6815

78. 32π 79. 16

380. π(1 − 1

e) 81. π

1282. 11

64π 83. π2

1284. 2511

3285. 45

32π 86.

(0, 24−15π

30π−20

)87. 16

388. π

89. 329|a|3 90. 2+π, 0 91.

(0, 24−15π

30π−20

)92.

(0,−1

4

)93. (0, 1) 94.

(3

π√2, 3π√2

)95. a

296.

(25, 25

)97.

(2532, 235112

)98. 320π

99. 315kπ 100. 7

3kπ 101. 344

75k 102. 3π+8


matematika ii (mathematics ii)

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