Indefinite Integral

Integration is used in dealing with two essentially di?erent types of problems:

The first types are problems in which the derivative of a function, or its rate of change, or the slope of its graph, is known and we want to find the function. We are therefore required to reverse the process of differentiation. This reverse process is known asanti-differentiation, or finding a primitive function or finding anindefinite integral.

The second type leads to the definition of thedefinite integral.Definite integrals are used for finding area, volume, centre of gravity, moment of inertia, work done by a force, and in many other applications.

This integral is called an indefinite integral because its value is not fully determined until the endpoints are specified. This ambiguity is dealt with by the addition of the constant C at the end. This constant of integration is added to the end because there are, in fact, an infinite number of solutions to the integral.From exam point of view the whole category of integrals are widely used in modern sciences as well as in engineering entrance examinations. They account to many lengthy numerical and key derivations.

Basic ConceptLet F(x) be a differentiable function of x such that d/dx [F(x)] = f(x). Then F(x) is called the integral of f(x). Symbolically, it is written asf(x) dx = F(x).f(x), the function to be integrated, is called the integrand.F(x) is also called theanti-derivative (or primitive function) of f(x).As the differential coefficient of a constant is zero, we have.d/dx (F(x)) = f(x) => d/dx [F(x)+c] = f(x).Therefore,f(x) dx = F(x) + c.This constant c is called the constant of integration and can take any real value.INTEGRATION AS THE INVERSE PROCESS OF DIFFERENTIATIONBasic formulae:Anti-derivatives or integrals of some of the widely used functions(integrands) are given below.Standard Formulae:Example - 1: Evaluate sin x / 1 + sin x dx.Solution: = sin x + 1 1 = 1 1/1+sin x 1sinx/1sin x = 1 1sinx / 1sin2x = 1 1 / cos2x sinx / cos2x = 1 sec2x sec x tan x Nowd/dx (x tan x sec x) = 1 sec2x sec x tan x(1 sec2 x sec x tan x) dx = x tan x sec x + c.

Example - 2:Evaluate sin3x+cos3x / sin2x cos2xdx.Solution:=tanx secx dx +cot xcosecx dx=secx cosecx + c.

METHODS OF INTEGRATIONIf the integrand is not a derivative of a known function, then the corresponding integrals cannot be found directly. In order to find the integral of complex problems, generally three rules of integration are used.(1)Integration by substitution or by change of the independent variable.(2)Integration by parts.(3)Integration by partial fractions.INTEGRATION BY SUBSTITUTIONDirect SubstitutionIf integral is of the form f(g(x)) g'(x) dx, then put g(x) = t, provided f(t)exists.(i). f'(x)/f(x) dx= ln |f (x)| + c Put f (x) = t=>f'(x) dx = dt=> dt/t= ln |t| + c = ln |f (x)| + c.(ii). f'(x)f(x) dx = 2f(x)+c Put f (x) = t=. dt/t = 2t + c = 2f(x) + c.Example -3:Evaluate sin(Inx) / x dx.Solution:Let lnx = t. Then dt = 1/xdx Hence I =sint dt = -cost + c = -cos(lnx) + cExample -4:Evaluate x3tan1x4/ 1 + x8dx.Solution:Put tan1x4= t=> 4x3/1+x8dx = dt I =1/4 t dt = 1/4 t2/ 2 + A = 1/8=(tan1x4)2+ A.Example -5:Evaluate (2x2+ 1) sin2(2x + 3x) dx.Solution:Letz = 2x33+ 3x dz =(6x2+3)dx = 3( 2x2+1)dx =1/3 sin2zdz = 1/3 1cos2z/2 dz =1/6 (1cos2z)dz = 1/6 [z sin2z / z] + c=1/6[2x2+ 3x sin(4x3+ 6x)/2] = + c

Standard Substitutions:For terms of the form x2+ a2orx2+ a2, put x = a tanora cotFor terms of the form x2- a2orx2 a2, put x = a secoracosec (A)For terms of the form a2- x2orx2+ a2, put x = a sinora cosIf botha+x,ax,are present, then put x = a cos.For the type(xa)(bx), put x = a cos2+ b sin2For the type(x2+a2x)nor(xx2a2)n, put the expression within thebracket = t.For the type(nN, n> 1), put x+b/x+a = tFor1/(x+a)n1(x+b)n2, n1,n2N(and > 1), again put (x + a) = t (x + b)Example -6:EvaluateSolution:= I2 Put 1 x = t2 dx = 2t dt I = 1x +x1x + sec11x + kExample -7:Evaluate. dx / (x+1)6/5(x3)4/5

Indirect SubstitutionIf the integrand is of the form f(x)g(x), whereg(x) is a function of the integral of f(x), then put integral of f(x) = t.Example -8:EvaluateSolution: Integral of the numerator= x3/2/ 3/2Put x3/2= t.We get I=2/3 dt/t2+a3=2/3 In |x3/2 +x3+a3|+ c.Derived Substitution:Some time it is useful to write the integral as a sum of two related integrals which can be evaluated by making suitable substitutions.Examples of such integrals are:A. Algebraic Twins

B. Trigonometric twins

Method of evaluating theseintegral are illustrated by mean of the followingexamples:Example -9:Evaluate 5/ 1+ x4dx.

ForI2,we writex +1/x t ==>(1 1/x2)dx = dtCombining the two integrals, we getExample -10:Evaluatetanx dx.Solution: Puttanx =t2=>sec2x dx = 2t dt =>dx = 2tdt / 1 +t4

This can be solved by the method used in example (9).

INTEGRATION BY PARTSIf u and v be two functions of x, then integral of product of these two functions is given by:Note:In applying the above rule care has to be taken in the selection of the first function(u) and the second function (v). Normally we use the following methods:If in the product of the two functions, one of the functions is not directly integrable(e.g. lnx, sin-1x, cos-1x, tan-1x etc.) then we take it as the first function and the remaining function is taken as the second function. e.g. In the integration ofx tan-1x dx, tan-1x is taken as the first function and x as the second function.If there is no other function, then unity is taken as the second function e.g. In the integration oftan-1x dx , tan-1x is taken as the first function and 1 as the second function.If both of the function are directly integrable then the first function is chosen in such a way that the derivative of the function thus obtained under integral sign is easily integrable. Usually we use the following preference order for the first function.(Inverse, Logarithmic, Algebraic, Trigonometric, Exponent)In the above stated order, the function on the left is always chosen as the first function. This rule is called asILATEe.g. In the integration of,x is taken as the first function and sinx is taken as the second function.Example -11: Evaluate.Solution:

Example -12:Evaluatesec3d.Solution: I =sec3d= secsec2d tan(sec tan) d =sectan (sec2 1)sec d= sectan I + ln|sec+ tan|=>I=1/2 [sec tan] + 1/2 In |sec tan| + cExample -13: Evaluate x3In xdx.Solution:Hereu = lnx=>du = 1/x dx and v = x3=>dv = 3x2dx I = 1/4 x4In x 1/4 x4In x 1/16 x4+ c =x4/ 4 . In x 1/4x3dx= 1/4x4In x 1/6x4+ c.An important result:In the integralg(x)exdx,if g(x) can be expressed as g(x) = f(x) + f'(x) then=ex[f(x) + f(x)] dx = exf(x) + cExample -13: Evaluateex(1/x 1/x2)dx.Solution:Let I =ex(1/x 1/x2)dx Here f(x) = 1/x, f'(x) = 1/x2 Hence I =ex/ x + c

Example -14: Evaluateex(1x / 1+x2)2dx.

INTEGRATION BY PARTIAL FRACTIONSA function of the form P(x)/Q(x), where P(x) and Q(x) arepolynomials, is called a rational function. Consider the rational function x+7 / (2x3)(3x+4) = 1/2x3 1/ 3x+4The two fractions on the RHS are called partial fractions. To integrate the rational function on the LHS, it is enough to integrate the two fractions on the RHS, which are easily integrable. This is known as method of partial fractions. In case the degree of P(x) (numerator)is not less than that of Q(x) (denominator), we carry out the division of P(x) by Q(x) and reduce the degree of the numerator.In order to write P(x)/Q(x) in partial fractions, first of all we writeQ(x) = (x a)k...(x2+x + )r... where binomials are different, and then setwhere A1, A2, ..., Ak, M1, M2, ......, Mr, N1,N2, ...... ,Nrare realconstants to be determined. These are determined by reducing both sides of the above identity to integral form and equating the coefficients of equal powers of x, which gives a system of linear equations in the coefficient. (This method is called the method of comparison of coefficients). The constants can also be obtained by substituting suitably chosen numerical values of x in both sides of the identity.Note:Before proceeding to write a rational function as a sum of partial fractions, we should be taken to ascertain that it is either a proper rational fraction or is rewritten as one.A rational functionP(x)/Q(x) is proper if the degree of polynomial Q(x) is greater than the degree of the polynomial P(x). In case the degree of P(x) greater than or equal to the degree of Q(x), we first write P(x)/Q(x)= h(x) + P(x)/Q(x), whereh(x) is a polynomial and p(x) is a polynomialof degree less than the degree of polynomial Q(x).Example -15: Evaluate cos x dx / (1+sinx)(2+sinx).Solution: Putsinx = t=>cosx dx= dt=log (1+ t) log (2 +t ) +c = log(1+sinx)/(2+sinx) + cExample -16: Evaluate 1/ sinx (2+cos x2 sinx) dx.Solution: Puttan= t=>sec2.x/2 . 1/2dx = dt=>dx = 2/sec2x/2 dt => 2/1 +t2dt, then we have Expands into simple fractions 1 +t2/ t(t3) = A/t + B/t3 + C/t1 After solve the coefficients, A = 1/3;B = 5/3;C = 1 HenceI = 1/3 dt / t + 5/3 dt/ t3 dt / t1

= 1/3 In |t| + 5/3 |t3| In |t1| + c =1/3 In |tanx/2| + 5/3|tan x/2 3| IN | tan x/2 1 | + c.Example -17:Evaluate xdx / x3+ 1.Solution:Since x3+ 1 = (x + 1) (x2- x + 1) (the second factor is not a product of linear factors), the partial fractions of the given integer will have the form. x/x3+ 1= A / x + 1 + Bx + D / x2 x + 1 Hence, x = A(x2- x + 1) + (Bx + D)(x + 1) = (A + B)x2+ (-A + B + D)x + (A + D). Equating the coefficients of equal powers of x, we get A = -1/3, B = 1/3, D = 1/3. Thus l = 1/3 dx/ x+1 + 1/3 x+1/x2x+1 dx = 1/3 |x+1| + 1/3 l1. To calculate the integrall1= x + 1 / x2 x + 1 dx. express (x + 1) = l(d.c. of x2- x + 1) + m=>(x + 1) = l(2x - 1) + m = 2xl - l + m Comparing the coefficients of like powers of x,we get l =1/2 and m=3/2=>l1= 1/2 2x1/x2x2 x+1 dx + 3/2 dx /x2 x + 1

Integrals of the Form:

In these type of integrals we write px + q =(diff coefficient of ax2+ bx + c) + m.Findand m by comparing the coefficient of x and constant term on both sides of the identity. In this way the question will reduce to the sum of two integrals which can be integrated easily.Integral of the typeIn this case substitute ax2+ bx + c = M (px2+ qx + r) + N (2px + q) + RFind M, N, & R. The integration reduces to integration of three independent functions.Example: Evaluate.Solution: Let x + 1 = (differential Coefficient of 2x2+ x 3) + B x +1 = A(4x + 1) + B = 4ax + A + B equating the coefficients, A =, B = Now I = = Let I1=and I2= Put 2x2+ x-3 =z (4x +1) dx =dz I1=== I2== = Hence I =.Integration of Irrational Algebraic Fractions:1. Irrational functions of (ax+b)1/n and x can be easily evaluated by the substitutiontn= ax+ b. Thus.2. . Here we substitute, x k = 1/t.This substitution will reduce the given integral to -.3. , so that.Now the substitution C + Dt2= u2reduces it to the form.4. Here, we write, ax2+bx +c = A1(dx +e) ( 2fx +g) +B1( dx +e) +C1where A1, B1and C1are constants which can be obtained by comparing the coefficient of like terms on both sides. And given integral will reduce to the formA1Example -20: Evaluate.Solution:Put x + 1 = t2, we getTrigonometric IntegralsIntegrals of the formR ( sinx, cosx) dx:Here R is a rational function of sin x and cos x. This can be translated into integrals of a rational function by the substitution: tan(x/2) = t. This is the so called universal substitution. In this case.Sometimes, instead of the substitution tanx/2 = t, it is more advantageous to make the substitution cot x/2 = tUniversal substitution often leads to very cumbersome calculations. Indicated below are those cases where the aim can be achieved with the aid of simpler substitutions.(a) If R(-sin x, cos x) = -R(sin x, cos x), substitute cos x = t(b) If R(sin x, -cos x) = -R(sin x, cos x), substitute sin x = t(c) If R(-sin x, - cos x) = R(sin x, cos x), substitute tan x = tIntegrals of the form:

Rule for (i) : In this integral express numerator as l (Denominator) + m(d.c. of denominator) + n. Find l, m, n by comparing the coefficients of sinx, cosx and constant term and split the integral into sum of three integrals.

Rule for (ii) : Express numerator as l (denominator) + m(d.c. of denominator) and find l and m as aboveExample : Evaluate.Solution:If in expressionwe substitute -sinx for sin x, then the integrand will change its sign. Hence, we take advantage of the substitutiont = cosx; dt = - sinx dx. This gives

TRIGONOMETRIC INTEGRALSIntegrals of the formR ( sinx, cosx) dx:Here R is a rational function of sin x and cos x. This can be translated into integrals of a rational function by the substitution:tan(x/2) = t. This is the so called universal substitution. In this casesinx = 2t/1+t2; cos x = 1t2/ 1+t2, x = 2 tan1t; dx = 2dt/1+t2Sometimes, instead of the substitution tanx/2 = t, it is more advantageous to make the substitution cot x/2 = tUniversal substitution often leads to very cumbersome calculations. Indicated below are those cases where the aim can be achieved with the aid of simpler substitutions.(a)If R(-sin x, cos x) = -R(sin x, cos x), substitute cos x = t(b)If R(sin x, -cos x) = -R(sin x, cos x), substitute sin x = t(c)If R(-sin x, - cos x) = R(sin x, cos x), substitute tan x = tIntegrals of the form:(i)pcosx + q sinx + r / a cos x + b sin x + c dx (ii)p cosx + q sinx / a cos x + b sin x dx

Rule for (i) : In this integral express numerator as l (Denominator) + m(d.c. of denominator) + n. Find l, m, n by comparing the coefficients of sinx, cosx and constant term and split the integral into sum of three integrals.ldx + md.c. of (Denominator) / denominator dx + ndx/ a cos x + b sin x + c

Rule for (ii) : Express numerator as l (denominator) + m(d.c. of denominator) and find l and m as aboveExample -21: Evaluatedx / sin x (2 cos2x1).Solution:If in expression 1/sinx(2cos2x1)we substitute -sinx for sin x, then the integrand will change its sign. Hence, we take advantage of the substitutiont = cosx; dt = - sinx dx. This givesIntegration of the Type:(sinMxcosNx)dx

M & Nnatural numbers.If one of them is odd, then substitute for term of even power.If both are odd, substitute either of the term.If both are even, use trigonometric identities only.Example -22: Evaluatesin x cos x dx.Solution:Putsinx= tt1/2= 2/3 t3/2+ c = 2/3 (sinx)3/2+ c.Example -23: Evaluatesin2x cos4x dx