Lecture 3

Techniques of integration (cont’d)

Integration by parts (cont’d)

Relevant section from Stewart, Eighth Edition: 7.1

In the previous lecture, we derived the following formula:

Integration by Parts: Formula 1

∫

u(x)v′(x) dx = u(x)v(x) −∫

u′(x)v(x) dx . (1)

This formula is often written in an alternative fashion, using the differentials of the functions involved

in the integrand, as opposed to functions multiplied by the differential “dx”. Here:

du = u′(x) dx and dv = v′(x) dx , (2)

so that Formula 1 may be rewritten as

Integration by Parts: Formula 2

∫

u dv = uv −∫

v du . (3)

Example 2: Let us revisit Example 1 from the previous lecture, i.e.,

∫

xex dx , (4)

and show how it is expressed in terms of Formula 2. First of all, as before, we’ll let

u(x) = x or simply u = x , (5)

which implies that

du = u′(x) dx = dx or simply du = dx . (6)

Secondly, we’ll let

dv = ex dx . (7)

But dv = v′(x) dx, which implies that

v′(x) = ex, (8)

22

which, in turn, implies that

v(x) = ex . (9)

Note that we did not write that

v(x) = ex + C , (10)

where C is an arbitrary constant. You don’t need C – we’ll show why later. Formula 2 then becomes

∫

u dv = uv −∫

v du∫

xex dx = xex −∫

exdx

= xex − ex + C . (11)

Let’s now see what happens if we included the arbitrary constant C in our expression for v(x) above:

∫

u dv = uv −∫

v du∫

xex dx = x[ex +C]−∫

[ex + C]dx

= xex + Cx− ex − Cx+D

= xex − ex +D , (12)

where D is an arbitrary constant. This is the same result as was obtained without the use of C. As

such, we see that using the arbitrary constant C inside the series of steps involved in integration by

parts is redundant. (This, of course, is the case when you use antiderivatives to evaluate a definite

integral using FTC II. You can add any constant you wish to the antiderivative to produce another

antiderivative. But when you take the difference of the antiderivatives evaluated at x = a and x = b,

this constant disappears.

Example 3: Find∫

x2ex dx . (13)

From the experience gained in Examples 1 and 2, it seems that letting u = x2 would be a good

idea. Integration by parts will then produce an integral with a lower power of x. As such, using the

differential notation of Formula 2,

u = x2 =⇒ du = 2x dx , (14)

23

and

dv = ex dx = v′ dx =⇒ v = ex . (15)

Using Formula 2,∫

u dv = uv −∫

v du∫

x2ex dx = x2ex −∫

ex2x dx

= x2ex − 2

∫

xex dx

= x2ex − 2(xex − ex) + C

= x2ex − 2xex + 2ex + C , (16)

where we have used the results of Examples 1 and 2. Of course, it is always good practice to check

the result:

d

dx

[

x2ex − 2xex + 2ex + C]

= 2xex + x2ex − 2ex − 2xex + 2ex

= x2ex , (17)

which verifies our result.

Example 4: Find∫

x sinx dx . (18)

Once again, it seems that it would be a good idea if we set u = x so that du = dx. Then

dv = sinx dx = v′(x) dx =⇒ v(x) = − cos x . (19)

Then, using Formula 2,∫

udv = uv −∫

vdu∫

x sinx dx = −x cos x−∫

(− cos x) dx

= −x cos x+

∫

cos x dx

= −x cos x+ sinx+ C . (20)

Once again, we verify the result,

d

dx[−x cos x+ sinx+ C] = − cos x+ x sinx+ cos x

= x sinx . (21)

24

Example 5: Find∫

lnx dx . (22)

At first sight, this might appear confusing, since there is only one function in the integrand. How can

we express it as a product. The trick is to express the integrand as

lnx = (1)(ln x) = (lnx)(1) . (23)

There are two choices for our integration by parts procedure. If we let u = 1, then du = 0. Further-

more, by letting dv = lnx dx, we would have to antidifferentiate lnx, which is what we’re trying to

do in the first place!

So, in what might appear to be quite contradictory to what we were doing earlier, let’s let

dv = 1 dx = v′(x) dx =⇒ v(x) = x , (24)

and

u = lnx =⇒ du =1

xdx . (25)

Now use Formula 2,

∫

udv = uv −∫

vdu∫

lnx = (lnx)(x)−∫

x · 1xdx

= x lnx−∫

dx

= x lnx− x+ C . (26)

Let’s verify the result:

d

dx[x lnx− x+ C] = lnx+ 1− 1

= lnx . (27)

Integration by parts applied to definite integrals

Let’s return to our original derivation of the integration by parts formula: Given two functions u and

v, we note thatd

dx[u(x)v(x)] = u′(x)v(x) + u(x)v′(x) , (28)

25

which implies (rather trivially, of course, but that’s not the point) that u(x)v(x) is the antiderivative

of the expression on the RHS. This implies that, by the use of the Fundamental Theorem of Calculus

II,∫ b

a[u′(x)v(x) + u(x)v′(x)] dx = u(x)v(x)

∣

∣

∣

b

a= u(b)v(b) − u(a)v(a) . (29)

We’ll rewrite the first equation as follows,

Definite integration by parts: Formula 1

∫ b

au(x)v′(x)dx = u(x)v(x)

∣

∣

∣

b

a−∫ b

au′(x)v(x) dx . (30)

This is integration by parts as applied to definite integrals. You’ll see that this is simply Formula 1

“evaluated at both limits x = a and x = b. It shouldn’t be too difficult to see that Formula 2 for

definite integrals becomes

Definite integration by parts: Formula 2

∫ b

au dv = uv

∣

∣

∣

b

a−∫ b

av du . (31)

Example 6: Find∫ e

1lnx dx . (32)

We’ll use two methods which, of course, are related to each other:

• Method No. 1: Let’s simply use the antiderivative of the integrand, as it was found in Example

5 above. From FTC II,

∫ e

1lnx dx = x lnx− x

∣

∣

∣

e

1

= (e ln e− e)− (1 ln 1− 1)

= e− e+ 1

= 1 . (33)

• Method No. 2: We’ll use the definite integration by parts method Formula 2:

u = lnx =⇒ du =1

xdx , (34)

26

and

dv = dx = v′(x) dx =⇒ v = x , (35)

so that

∫ e

1lnx dx = x lnx

∣

∣

∣

e

1−∫ e

1dx

= e ln e− 1 ln 1− x∣

∣

∣

e

1

= e− 0− (e− 1)

= 1 , (36)

in agreement with Method No. 1.

Example 7: (not done in class) Find∫ π

0x sinx dx . (37)

Here, we’ll let

u = x =⇒ du = dx ,

and

dv = sinx dx = v′(x) dx =⇒ v = − cos x , (38)

so that

∫ πx

0sinx dx = −x cosx

∣

∣

∣

π

0−∫ π

0(− cos x) dx

= −π cos(π)− (0 cos 0 ) +

∫ π

0cos x dx

= π + sinx∣

∣

∣

π

0

= π . (39)

An interesting family of definite integrals which are connected by recursion

Let’s consider the following family of antiderivatives,

∫

sinn x dx , n ≥ 0 . (40)

Of course, the first couple of antiderivatives are easy,

∫

sin0 x dx =

∫

dx = x+ C

∫

sinx dx = − cos x+ C . (41)

27

But at n = 2, the following antiderivative,

∫

sin2 x dx , (42)

is not obvious. Actually, we’ll be dealing with trigonometric integrals in the next lecture. Here we

wish to show that integration by parts can be used to establish a relation between antiderivatives with

different n values.

Let’s express sinn x as a product that can be treated with integration by parts, i.e.,

∫

sinn x dx =

∫

sinn−1 x sinx, dx . (43)

We’ll now let

u = sinn−1 x =⇒ du = (n− 1) sinn−2 x cos x dx , (44)

so that integration by parts will decrease the exponent of sinx. This implies that and

dv = sinx dx =⇒ v = − cos x . (45)

Now use IP (integration by parts) Formula 2:

∫

u dv = uv −∫

v du∫

sinn x dx = − sinn−1 x cos x−∫

(− cos x)(n − 1) sinn−2 x cos x dx (46)

= − sinn−1 x cos x+ (n− 1)

∫

sinn−2 cos2 x dx.

We’ll now rewrite the cos2 x in the integrand as cos2 x = 1− sin2 x, to obtain the equation,

∫

sinn x dx = − sinn−1 x cos x+ (n− 1)

∫

sinn−2 x dx− (n− 1)

∫

sinn x dx . (47)

Now notice that the integral in the last term on the right is the same as the integral on the left side,

so we’ll bring it over to give

n

∫

sinn x dx = − sinn−1 x cos x+ (n− 1)

∫

sinn−2 x dx . (48)

Finally, we’ll divide by n to obtain,

∫

sinn x dx = − 1

nsinn−1 x cos x+

(n − 1)

n

∫

sinn−2 x dx , n ≥ 2 . (49)

28

This is a kind of recursion equation that relates the antiderivatives of powers of sinx. (It is also given

in Stewart, Eighth Edition, as Example 6 on Page 475 (and derived on Page 476).

We now show how this relation is useful to compute values of the following general family of

definite integrals,

In =

∫ π/2

0sinn x dx , n ≥ 0 . (50)

First of all, let’s compute the first two integrals,

I0 =

∫ π/2

0dx =

π

2, (51)

and

I1 =

∫ π/2

0sinx dx

= (− cos x)∣

∣

∣

π/2

0

= (−0)− (−1)

= 1 . (52)

We’ll now use the definite integral form of the recursion relation in (49),

∫ π/2

0sinn x dx = − 1

nsinn−1 x cos x

∣

∣

∣

π/2

0+

(n− 1)

n

∫ π/2

0sinn−2 x dx , n ≥ 2 . (53)

For n ≥ 2,

sinn−1 0 = 0 and cosπ

2= 0 , (54)

so that the first term on the right side of the equation vanishes. As a result, we have

∫ π/2

0sinn x dx =

(n− 1)

n

∫ π/2

0sinn−2 x dx , n ≥ 2 , (55)

or simply,

In =n− 1

nIn−2 , n ≥ 2 . (56)

This is a recursion relation that relates In to In−2. From I0, we can determine I2, I4, etc.. From I1,

we can determine I3, I5, etc.. Let’s look at the first couple of terms of the even-indexed series,

I2 =1

2I0 =

1

2· π2

I4 =3

4I2 =

3

4· 12· π2. (57)

29

In general, we have

I2n =1 · 3 · 5 · · · (2n − 1)

2 · 4 · 6 · · · (2n) · π2. (58)

Now examine the first couple of terms of the odd-indexed series,

I3 =2

3I1 =

2

3

I5 =4

5I3 =

4

5· 23. (59)

In general, we have

I2n+1 =2 · 4 · 6 · · · (2n)

3 · 5 · 7 · · · (2n+ 1). (60)

30

Appendix: Some material covered in the Monday Section 005 Tuto-

rial

The definite integral and its applications (cont’d)

Both of these topics are particular examples of the “Spirit of Calculus” at work. They are taken from

ERV’s lecture notes for MATH 137P Fall 2012 (Week 12).

The work done by a nonconstant force

We start with a result that is well-known to you from high school physics. Suppose that a constant

force F = F i acts on a mass m, causing it to move along the x-axis from position x = a to x = b.

Then the total work done W by the force is given by the product of the magnitude of the force and

the displacement of the mass, i.e.,

W = F (b− a) (61)

This is a special case of the more general result in which a constant force F moves the mass in a

straight line that is not necessarily parallel to F. If the displacement vector of the mass is d, then the

total work W done by F is

W = F · d. (62)

In the discussion that follows, it will be sufficient to consider Eq. (61).

Now suppose that the force F is no longer constant, i.e., F = f(x)i, where the function f(x) is

not necessarily constant. If the mass m is moved from position x = a to x = b, what is the total work

W done by the force? You have most probably seen the answer in your first-year Physics course. It

is given by the definite integral,

W =

∫ b

af(x) dx. (63)

We now derive this result mathematically in terms of our Riemann sum definition of the definite

integral. And our derivation will be done by employing what we have previously called the “Spirit

of Calculus.” Very briefly, we’ll subdivide the interval [a, b] into tiny subintervals Ik of length ∆x,

and then approximate the force function f(x) as a constant over each subinterval. We then use the

constant-force result from Eq. (61) over each subinterval Ik, to approximate the work ∆Wk done in

moving the mass over the subinterval Ik. Finally, we sum over the contributions from all subintervals.

31

As before, we first consider an n > 0 (with the idea of letting n → ∞) and define

∆x =b− a

n. (64)

Then define the partition points,

xk = a+ k∆x, k = 0, 1, 2, · · · . (65)

Note that x0 = a and xn = b. These partition points define a set of n subintervals Ik = [xk−1, xk],

k = 1, 2, · · · n, of equal length ∆x.

Now select a sample point x∗k ∈ [xk−1, xk] from each subinterval Ik. Then evaluate the force

function f at each sample point x∗k. We now consider each value f(x∗k) as the approximation of f(x)

over the subinterval Ik. In other words, the function f(x) is approximated by a constant function

f(x∗k). In this way, we may use Eq. (61) to approximate the work ∆Wk done by the function f(x) in

moving the mass over the subinterval Ik, i.e., from xk−1 to xk, as follows,

∆Wk∼= f(x∗k)∆x (constant force strength × displacement). (66)

The total work done by the force in moving the mass from x0 = a to xn = b will then be approximated

as follows,

W =

n∑

k=1

∆Wk∼=

n∑

k=1

f(x∗k)∆x. (67)

But by construction, the RHS of this equation is a Riemann sum for the definite integral of f(x) from

x = a to x = b. Assuming that the definite integral of f exists (which is ensured if f is continuous or

piecewise continuous), we have

W = limn→∞

n∑

k=1

f(x∗k)∆x

=

∫ b

af(x) dx. (68)

This concludes our mathematical justification of the definite integral formula for work.

An important note regarding the dimensions of work and the integral formula

The dimensionality of force isMLT−2. (Think of F = ma and the dimensions of mass and acceleration.

Therefore the dimensionality of work is force times distance, or ML2T−2. Note that the dimensionality

32

of the Riemann sum in Eq. (67) is also force times distance. Since the definite integral in Eq. (68)

is the limit of Riemann sums with this dimensionality, it follows that the definite integral has the

dimension of work. Basically, we can think of the integrand as having the dimensionality of force and

the infinitesimal dx as the dimensionality of length.

The extension of the work integral to several dimensions and motion along curves

In a future course on advanced calculus that includes the subject of “vector calculus” (e.g., AMATH

231 or MATH 227), you will consider the more general case of a nonconstant force F(r) in R3 acting

on a mass m as the mass moves along a curve C from a point P to point Q. The situation is sketched

in the diagram below.

xy

z

P

Q

m F(r(t))r(t)

The goal is once again to compute the total amount of work W done by the force. Once again, in

the “Spirit of Calculus,” the idea is to break up the motion into tiny pieces over which we can use the

constant-force-straight-line formula “W = F (b − a)” to approximate the work over these pieces. We

then “sum up,” i.e., integrate over all contributions to obtain W . In this case, since the force vectors

F(r) will not, in general, be parallel to the motion of the mass, we’ll have to take scalar products

of these vectors F with the instantaneous direction of motion of the mass m – in other words, the

velocity vectors v = r′ along the curve. The net result is that we have an integral of the following

form∫

CF · dr, (69)

which is known as the line integral of the vector field F over the curve C. You may already have

seen this integral in your Physics course.

33

The definite integral and its applications (cont’d)

The average value of a function

Suppose that a thin, straight wire is located on the x-axis, specifically on the interval [a, b]. Further-

more, suppose that the function f(x) represents the temperature of the wire at a point x ∈ [a, b]. The

question is, “What is the average temperature of the wire?” This is a specific example of the more

general question: What is the average value of a function f over the interval [a, b]?

We’ll address this problem in the usual way, i.e., by means of the “Spirit of Calculus.” We’ll divide

up the interval [a, b] into n subintervals Ik, take samples of the function f(x) on these subintervals,

and then compute the average of these sample values. The average value of f over the interval [a, b]

will be the limit n → ∞ of these average values, provided that the limit exists.

So, as before, let n > 0 and define ∆x =b− a

n. Then define the partition points,

xk = a+ k∆x, k = 0, 1, · · · , n, (70)

so that x0 = a and xn = b. These points define the n subintervals Ik = [xk−1, xk], k = 1, 2, · · · , n.From each subinterval Ik, choose a sample point x∗k ∈ Ik. Then evaluate the function at this sample

point. The result is a set of n function values f(x∗k). These may be viewed as samples of the function

f(x) over the interval [a, b].

It seems reasonable to take the average of these n function values – we’ll denote this average as

f̄n =1

n

n∑

k=1

f(x∗k). (71)

Now the sum on the RHS looks almost like a Riemann sum to the definite integral of f . However, a

∆x is missing. So let’s multiply and divide by ∆x as follows,

f̄n =1

n

1

∆x

n∑

k=1

f(x∗k)∆x

=1

n∆xSn. (72)

Here, Sn is a Riemann sum corresponding to the definite integral

∫ b

af(x) dx. There remains the

question about what to do about the factor1

n∆x. Recalling the definition of ∆x:

∆x =b− a

n⇒ n∆x = b− a. (73)

34

Therefore, the average value in (72) becomes

f̄n =1

b− aSn. (74)

Assuming that f is continuous (or at least piecewise continuous), the limit of the Riemann sums Sn

exists, and we have

limn→∞

f̄n =1

b− alimn→∞

Sn =1

b− a

∫ b

af(x) dx. (75)

This is the average value of f over the interval [a, b], which we shall denote as follows,

f̄[a,b] =1

b− a

∫ b

af(x) dx. (76)

In other words, we compute the definite integral of f over the interval [a, b] and then divide by the

length of the interval, b− a.

Let’s now rewrite Eq. (76) as follows,

∫ b

af(x) dx = f̄[a,b](b− a). (77)

If we assume, for the moment - for the sake of simplicity - that f(x) > 0 on [a, b], then Eq. (77) is

stating that the area enclosed by the graph of f(x), the lines x = a and x = b and the x−axis is given

by the average value of f on [a, b] multiplied by the length of the interval (b− a). In other words, as

sketched in the figure below, we have replaced the area enclosed by the graph, etc., by a rectangle of

height f̄[a,b]. The rectangular region is shaded.

a bx

y = f(x)

f̄[a,b]

That being said, we may now relax the restriction that f(x) be strictly positive on [a, b]. In this case,

the average value of f on [a, b] times the length (b− a) will be a signed area.

35

Some simple, yet illuminating, examples:

(Note: These may not have been covered in the tutorial, but are added for your information.)

1. The function f(x) = 1 over the interval [a, b] = [0, 1]. Since f(x) assumes only one value over

the entire interval, namely, the value 1, we expect that its average value is 1. Let’s check this.

Since a = 0, b = 1 and f(x) = 1, we have

f̄[0,1] =1

1

∫ 1

01 dx = [x]10 = 1, (78)

as expected.

2. The function f(x) = x over the interval [a, b] = [0, 1]. From a look at the graph of f over

[0, 1], we might guess that the average value is its average value is 1/2. Since a = 0, b = 0 and

f(x) = 1, we have

f̄[0,1] =1

1

∫ 1

0x dx =

[

x2

2

]1

0

=1

2. (79)

Our intuition was correct.

3. The function f(x) = x2 over the interval [a, b] = [0, 1]. A look at the graph of f(x) = x2 shows

that there are many more x-values for which f(x) < 1/2 than in the previous case, f(x) = x.

Therefore, we would expect the average value to be less than 1/2. Since a = 0, b = 0 and

f(x) = 1, we have

f̄[0,1] =1

1

∫ 1

0x2 dx =

[

x3

3

]1

0

=1

3. (80)

4. In general, the function f(x) = xn over the interval [a, b] = [0, 1], where n > 0. Since a = 0,

b = 0 and f(x) = 1, we have

f̄[0,1] =1

1

∫ 1

0xn dx =

[

xn+1

n+ 1

]1

0

=1

n+ 1. (81)

Note that as n → ∞, the average value of the function xn behaves as follows,1

n+ 1→ 0. Does

this make sense? For any x such that 0 ≤ x < 1, raising it to higher powers makes it smaller,

i.e., xn → 0 as n → ∞. (Think of x = 1/2.) That means that the graph of f(x) = xn gets flatter

and flatter as n increases, except at x = 1, since 1n = 1 always. This is illustrated below.

Since all values of xn for x ∈ [0, 1) – note that we exclude the case x = 1 – approach zero as

n → ∞, we expect the average value of xn to approach zero in the limit n → ∞.

36

An important piece of advice: When you encounter a new concept in mathematics, it is often

most helpful to apply that concept to a set of cases, perhaps a one-parameter family of functions, and

to observe the behaviour of the results as you vary the parameter. We have done this with the concept

of the average value of a function, applying it to the one-parameter family of functions xn on [0, 1].

37

Lecture 4

Trigonometric integrals

Relevant section from Stewart, Eighth Edition: 7.2

In this section we consider the problem of finding antiderivatives of intgrands that are products of

trigonometric functions. These integrals are actually encountered when the method of trigonometric

substitution, to be discussed the next lecture, is used.

Integrands involving powers of sin and cos

Here we consider integrals of the form

∫

sinm x cosn x dx . (82)

The simplest cases are well known to us,

∫

sinx dx = − cos x+ C

∫

cos x dx = sinx+ C . (83)

Example 1: (This case was not done in class.) The next simplest case is

∫

sinx cos x dx =

∫

u du (u = sinx, du = cos x dx)

=1

2u2 + C

=1

2sin2 x+ C . (84)

Note that we could also have performed the substitution u = cos x, du = − sinx dx,

∫

sinx cos x dx = −∫

u du

= −1

2u2 + C

= −1

2cos2 x+ C . (85)

At first glance, this result is different from the first result, but it is equivalent since

−1

2cos2 x+ C = −1

2(1− sin2 x) + C

=1

2sin2 x+D , (86)

38

where D is an arbitrary constant.

Note that from the double angle formula,

sin 2x = 2 sinx cos x , (87)

the above integral can also be evaluated as follows,

∫

sinx cos x dx =1

2

∫

2 sin x cos x dx

=1

2

∫

sin 2x dx

= −1

4cos 2x + C . (88)

This result appears to be quite different from the earlier two results. But with the double angle for-

mula for the cosine function (below), it can be shown that this result is equivalent to the other two.

We’ll leave this as an exercise for the reader.

Example 2: The next simplest cases are

∫

sin2 x dx

∫

cos2 x dx . (89)

In these cases, the following double angle formula for the cosine is useful:

cos 2x = cos2 x− sin2 x . (90)

If we rewrite the above formula as follows,

cos2 x = (1− sin2 x)− sin2 x

= 1− sin2 x , (91)

we then obtain the result,

sin2 x =1

2− 1

2cos 2x . (92)

We may then use this result to obtain,

∫

sin2 x dx =

∫[

1

2− 1

2cos 2x

]

dx

=1

2x− 1

4sin 2x+ C . (93)

39

We can use the following double angle formula for the sine function,

sin 2x = 2 sinx cos x , (94)

to rewrite the above result as∫

sin2 x dx =1

2x− 1

2sinx cos x+ C . (95)

If we rewrite the formula in (90) as follows,

cosx = cos2 x− (1− cos2 x)

= 2 cos2 x− 1 , (96)

we obtain

cos2 x =1

2+

1

2cos 2x . (97)

We then use this result to obtain∫

cos2 x dx =

∫[

1

2+

1

2sin 2x

]

dx

=1

2x+

1

4sin 2x+ C , (98)

which can also be written as∫

cos2 x dx =1

2x+

1

2sinx cos x+ C . (99)

Example 3(a): Let us now consider the following integral,∫

cos3 x dx . (100)

We’ll express the integrand as the following product,

cos3 x = cos2 x cos x (101)

to perform the following steps,∫

cos3 x dx =

∫

cos2 x cos x dx

=

∫

(1− sin2 x) cos x dx

=

∫

cosx dx−∫

sin2 cos x dx

= sinx− 1

3sin3 x+ C . (102)

40

The final integral was obtained by the method of substitution: Let u = sinx so that du = cos x dx.

Then

∫

sin2 cos x dx =

∫

u2 du

=1

3u3 + C

=1

3sin3 x+ C . (103)

Note that the integral in the second line of the original derivation could also be treated in one line

using substitution as follows: Letting u = sinx so that du = cos x dx,

∫

(1− sin2 x) cos x dx =

∫

(1− u2) du

= u− 1

3u3 + C

= sinx− 1

3sin3 +C , (104)

in agreement with the previous result. This is often the way that solutions are presented in Stewart’s

text.

Example 3(b): We can evaluate the following antiderivative,

∫

sin3 x dx , (105)

in a similar manner:

∫

sin3 x dx =

∫

sin2 sinx dx

=

∫

(1− cos2 x) sinx dx

=

∫

sinx dx−∫

cos2 x sinx dx

= − cos x+1

3cos3 x+ C . (106)

The above two examples are special cases of a more general method of treating an integral of the form

∫

sinm x cosn x dx . (107)

41

In the case that one of m or n is odd, then one tries, using the relation

sin2 x+ cos2 x = 1 , (108)

to express the integral in one of the following forms,∫

[ polynomial in cos x] sinx dx (109)

which can be treated by the substitution u = cos x, du = − sinx dx, or or∫

[ polynomial in sinx] cos x dx , (110)

which can be treated by the substitution u = sinx, du = cos x dx. A summary of the entire procedure

is given in the box with heading, “Strategy for Evaluating∫

sinm x cosn x dx,” on Page 481 of

Stewart’s textbook (Eighth Edition).

Example 3(a) revisited: Let us now reconsider the following integral,∫

cos3 dx =

∫

cos2 cos x dx . (111)

This time, however, instead of expressing cos2 x − (1 − sin2 x) as before, we’ll try the method of

integration by parts:

u = cos2 x =⇒ du = −2 cos x sinx dx , (112)

dv = cos x ex =⇒ v = sinx . (113)

Now use Formula 2 of Integration by Parts,∫

u dv = uv −∫

v du , (114)

to obtain∫

cos3 x dx =

∫

cos2 x cos x dx

= cos2 x sinx+ 2

∫

sin2 x cos x dx

= cos2 x sinx+2

3sin3 x+ C . (115)

At first glance, this does not look like the answer obtained in our first treatment of this integral, Eq.

(102). With a little rewriting, however, we obtain the same result, i.e.,

cos2 x sinx+2

3sin3 x+ C = [1− sin2 x] sinx+

2

3sin3 +C

= sinx− 1

3sin3 x+C . (116)

42

The “moral” of the story: Sometimes, you may obtain a result that does not agree with the result

presented in a book or table of integrals. This does not necessarily mean that your result is incorrect

- it may simply be the correct result, but cast in a slightly different form.

Example 4: Find∫

sin5 x cos2 x dx . (117)

If we substitute cos2 x = 1− sin2 x, we then obtain the following sum,

∫

sin5 x dx−∫

sin7 x dx . (118)

In other words, we are in a deeper mess! Instead, we’ll do the following,

sin5 x cos2 x = sin4 x cos2 x sinx

= (1− cos2 x)2 cos2 x sinx

= (1− 2 cos2 x+ cos4 x) cos2 x sinx

= (cos2 x− 2 cos4 x+ cos6 x) sinx . (119)

This is in keeping with the “Strategy” from Stewart’s textbook, where the power of the sine function,

5, is an odd number. As a result, our integrand is now in the form,

[ polynomial in sinx] cos x . (120)

We may now use the substitution method,

∫

sin5 x cos2 x dx =

∫

(cos2 x− 2 cos4 x+ cos6 x) sinx dx

= −∫

(u2 − 2u4 + u6) du (u = cos x, du = − sinx dx)

= −1

3u3 +

2

5u5 − 1

7u7 +C

= −1

3cos3 x+

2

5cos5 x− 1

7cos7 x+ C (resubstituing u = cos x) . (121)

Integrals involving powers of tan and sec

Here we consider integrals of the form

∫

tanm x secn x dx . (122)

43

Once again, these integrals are often encountered when we employ the method of trigonometric sub-

stitution.

First of all, let’s review the basic differentiation results of tan and sec:

d

dxtan x = sec2 x

d

dxsec x = secx tan x . (123)

If you have these results memorized, then fine. If not, they can always be derived quickly from first

principles:

d

dxtan x =

d

dx

[

sinx

cos x

]

=cos2 x+ sin2 x

cos2 x(quotient rule)

=1

cosx

= sec2 x . (124)

d

dxsec x =

d

dx

[

1

cosx

]

= − 1

cos2 x· (− sinx)

=1

cos x· sinxcos x

= sec x tan x . (125)

Also useful will be the following relationship between tan x and secx,

tan2 x = sec2 x− 1 or sec2 x− tan2 x = 1 or tan2 x+ 1 = sec2 x . (126)

Once again, if you have one or more of thse results memorized, fine. But if not, they can easily be

derived from the Pythagorean relation,

sin2 x+ cos2 x = 1 . (127)

Divide both sides by cos2 x:sin2 x

cos2 x+ 1 =

1

cos2 x, (128)

44

which, of course, is

tan2 x+ 1 = sec2 x . (129)

Example No. 1: The simplest integrals in this class, which you may have seen in MATH 137, are

∫

tan x dx = ln | sec x|+ C (130)

and∫

secx dx = ln | sec x+ tan x|+ C . (131)

Let’s derive these results. The first one is quite straightforward:

∫

tanx dx =

∫

sinx

cos xdx

= −∫

1

udu (u = cos x du = − sinx dx)

= − ln | cos x|+ C

= ln | cos x|−1 + C

= ln |(cos x)−1|+ C

= ln | sec x|+ C . (132)

The second result is obtained by means of a “clever trick:”

∫

sec x dx =

∫

sec x

(

sec x+ tan x

sec x+ tan x

)

dx (multiplication by 1)

=

∫

sec2 +secx tan x

sec x+ tanxdx

=

∫

1

udu (u = [sec x+ tan x] )

= ln |u|+ C

= ln | sec x+ tanx|+ C . (133)

(As mentioned in the lecture, some people would refer to this trick as “cheesy.” But it does achieve

the correct result.)

Example No. 2: The next simplest integral involving tan and sec is quite easy,

∫

sec x tanx dx = sec x+ C . (134)

45

Example No. 2: Let us now consider the integrals,

∫

tan2 x dx and

∫

sec2 x dx . (135)

These are rather straightforward:

∫

tan2 x dx =

∫

(sec2 −x) dx

= tan x− x+ C (136)

and

∫

sec2 x dx = tan x+ C . (137)

For higher powers of tan and sec, there is a strategy similar to that for powers of tan and sec: See the

box entitled, Strategy for evaluating∫

tanm x secm x. This strategy does not cover all cases and,

as the text indicates, one “may need to use identities, integration by parts, and occasionally a little

ingenuity. We consider a few examples below.

Example No. 3: The antiderivative,∫

tan3 x dx . (138)

Let’s write tan3 x = tan2 x tan x and then convert the tan2 x term:

∫

tan3 x dx =

∫

tan2 x tanx dx

= (sec2−1) tan x dx

=

∫

tanx sec2 x dx−∫

tan x dx

=1

2tan2 x− ln | sec x|+ C . (139)

The first integral was obtained from the method of substitution: u = tan x, du = sec2 x dx.

Example No. 4: The antiderivative,∫

sec3 dx . (140)

Method No. 1: (This method was started in class but not finished.) We’ll write sec3 x = sec2 x sec x

46

and then convert the sec2 x term:

∫

sec3 x dx =

∫

sec2 x sec x dx (141)

=

∫

(tan2 x+ 1) sec x dx (142)

=

∫

tan2 x sec x dx+

∫

sec x dx . (143)

The second integral is easy – we computed it earlier,

∫

secx dx = ln | sec x+ tan x|+ C . (144)

As for the first integral, we’ll try integration by parts,

∫

tan2 x sec x dx =

∫

tanx(sec x tanx) dx (145)

with u = tan x and du = sec x tan x dx, implying that v = sec x and du = sec2 x. Recalling that

∫

u dv = uv −∫

v du , (146)

we have∫

tan x(sec x tan x) dx = tanx sec x−∫

sec3 x dx . (147)

If we now substitute the results of (144 and (147) into (141), we obtain

∫

sec3 x dx = sec x tan x−∫

sec3 x dx+ ln | sec x+ tanx|+ C . (148)

It might appear that we have “goofed,” since the same integral appears on both sides of the equation!

But the saving grace is that the integral on the right side is multiplied by (-1). As such, we can bring

it over to the left side to give

2

∫

sec3 x dx = sec x tanx+ ln | sec x+ tanx|+ C . (149)

and then divide by 2 to arrive at the final result,

∫

sec3 x dx =1

2secx tan x+

1

2ln | sec x+ tanx|+ C . (150)

Method No. 2: (This was the method shown in class.) We’ll once again express the integrand as

sec3 x = sec2 x sec2 x but this time, instead of using a trigonometric identity to re-express the sec2 x

function, we’ll use integration by parts:

∫

sec3 x dx =

∫

sec2 x sec x dx . (151)

47

Let u = secx and dv = sec2 x dx so that du = sec x tanx dx and v = tanx. Then from the integration

by parts formula,∫

u dv = uv −∫

v du , (152)

we obtain the following,

∫

sec3 x dx = sec x tanx−∫

tanx sec x tan x dx . (153)

This result is actually equivalent to Eq. (147). We’ll work on the second integral by combining the

two tanx functions and using the trig identity:

∫

tan2 x sec x dx =

∫

(sec2 x− 1) sec x dx

=

∫

sec3 x dx−∫

sec x dx

=

∫

sec3 x dx− ln | sec x+ tanx|+ C . (154)

Inserting this result into (153) yields

∫

sec3 x dx = sec x tan x−∫

sec3 dx+ ln | sec x+ tanx|+C , (155)

which is identical to Eq. (148) from Method No. 1. As such, we shall be able to arrive at the same

result, i.e.,∫

sec3 x dx =1

2secx tan x+

1

2ln | sec x+ tanx|+ C . (156)

48

Lecture 5

Trigonometric substitution

Relevant section from Stewart, Eighth Edition: 7.3

In science and engineering, we often encounter integrals with integrands composed of square roots,

or integer powers of these square roots, of quadratic functions, e.g.,

√

A− x2,√

A+ x2,√

x2 −A . (157)

Such integrals can often be handled by means of appropriate “trigonometric substitutions,” the sub-

ject of this section. As we’ll see below, the goal is to remove the square root terms from

the integrands and this can be accomplished using appropriate trigonometric substitutions. These

substitutions will be based upon the following two important trigonometric relations,

sin2 x+ cos2 x = 1 ,

tan2 x+ 1 = sec2 x . (158)

As we discussed earlier, the second relation can be obtained from the first one by dividing both sides

of the first one by cos2 x. We’ll see that it is convenient to replace the constants A in the above

expresssions with a2, i.e.,√

a2 − x2,√

a2 + x2,√

x2 − a2 . (159)

Integrals which contain√a2 − x

2

Example 1: The antiderivative,∫

√

a2 − x2 dx . (160)

Let’s consider the change of variable,

x = a sin θ =⇒ dx = a cos θ dθ . (161)

We would like this to be a 1− 1 relationship between x and θ, which implies that the angle θ should

be restricted to the interval [−π/2, π/2]. More on this later.

49

In addition, we’ll assume, for simplicity, that a > 0. With this change of variable, the square root

term becomes,

√

a2 − x2 =√

a2 − a2 sin2 θ

= a√

1− sin2 θ

= a cos θ . (162)

Technically, the last line should be√

a2 − x2 = a| cos θ| , (163)

since the√· function is, by definition, non-negative. If we restrict the range of the θ variable to

[−π/2, π/2], then cos θ ≥ 0 and the absolute value sign can be omitted.

Inserting the expressions for the square root function and the dx term into the original integral,

we obtain

∫

√

a2 − x2 =

∫

[a cos θ][a cos θ] dθ

= a2∫

cos2 θ dθ . (164)

As you see, we’ve produced a trigonometric integral, the subject of the previous lecture. You may

recall that we can easily evaluate this integral using the double-angle formula for the cosine function,

∫

cos2 θdθ =

∫[

1

2+

1

2cos 2θ

]

dθ

=1

2θ +

1

4sin 2θ

=1

2θ +

1

2sin θ cos θ + C . (165)

But we must now express this result in terms of the original variable x. How do we do this? Let us

recall that

x = a sin θ =⇒ sin θ =x

a, (166)

which, in turn, implies that

θ = Sin−1(x

a

)

. (167)

This takes care of θ and sin θ. But what about cos θ?

One way to obtain cos θ is to simply use the trigonometric relation,

cos2 θ = 1− sin2 θ =⇒ cos2 θ = 1− x2

a2. (168)

50

This, in turn, implies that

cos θ =

√

1− x2

a2=

√a2 − x2

a. (169)

Since θ ∈ [−π/2, π/2], we can take the positive square root.

The above procedure of translating expressions involving θ to those involving x is simplified if we draw

a diagram that is associated with the change of variable,

x = a sin θ =⇒ sin θ =x

a. (170)

a

θ√

a2− x2

x

Note: This particular instructor thinks that it is always a good idea to draw the diagram.

We now combine all of our results to obtain the desired antiderivative,∫

√

a2 − x2 dx = a2∫

cos2 θ

=a2

2θ +

a2

2sin θ cos θ + C

=a2

2Sin−1

(x

a

)

+a2

2

(x

a

)

√a2 − x2

a+ C

=a2

2Sin−1

(x

a

)

+x

2

√

a2 − x2 + C . (171)

It’s a good idea to check this result by differentiation (omitting the arbitrary constant C):

d

dx

[

a2

2Sin−1

(x

a

)

+x

2

√

a2 − x2]

=a2

2

1√

1− x2

a2

(

1

a

)

+1

2

√

a2 − x2 +x

2· 12

1√a2 − x2

(−2x)

=1

2

[

a2√a2 − x2

+√

a2 − x2 − x2√a2 − x2

]

=1

2

[

a2 − x2√a2 − x2

+√

a2 − x2]

=√

a2 − x2 . (172)

Example 1 revisited - An application: Let us now apply this result to the following definite

integral,∫ a

−a

√

a2 − x2 dx , (173)

51

which is the area of the semi-circular region enclosed by the curve y =√a2 − x2 and the x-axis

between x = −a and x = b. This is one-half the area of a circle of radius a, namely1

2πa2.

From the FTC II and our earlier result,

∫ a

−a

√

a2 − x2 dx =a2

2Sin−1

(x

a

)

+x

2

√

a2 − x2

∣

∣

∣

∣

∣

a

−a

=

[

a2

2Sin−1(1) +

a

2

√

a2 − x2]

−[

a2

2Sin−1(−1) +

−a

2

√

a2 − x2]

=a2

2

(π

2

)

− a2

2

(

−π

2

)

=1

2πa2 . (174)

The above result was made possible by the fact that we had previously done all of the work to compute

the antiderivative of√a2 − x2. If, on the other hand, you were given the definite integral to evaluate,

and then started to employ the trigonometric substitution method, some time and work could be saved

by stopping after obtaining the antiderivative in terms of θ and evaluating it at appropriate limits, as

opposed to expressing your results in terms of x and then evaluating them. We illustrate this with

the problem addressed above.

First of all, note that the lower and upper limits of integration are −a and a, respectively. With

the change of variable,

x = a sin θ , (175)

this implies that the lower and upper limits of integration are −π/2 and π/2. Then,

∫ a

−a

√

a2 − x2 dx = a2∫ π/2

−π/2cos2 θ dθ

= a2[

θ

2+

1

4sin 2θ

]π/2

−π/2

= a2[π

2+ 0]

− a2[

−π

2+ 0]

=1

2πa2 , (176)

in agreement with our earlier result.

Example 2: The antiderivative,∫

1√a2 − x2

dx . (177)

52

(In the case that a = 1, you may remember that the antiderivative is Sin−1(x).) Once again, we

consider the change of variable,

x = a sin θ =⇒ dx = a cos θ dθ . (178)

Then, as before,√

a2 − x2 = a cos θ . (179)

Inserting these expressions into the integral,

∫

1√a2 − x2

dx =

∫

a cos θ

a cos θdθ

=

∫

dθ

= θ + C . (180)

But from the change of variable,

x = a sin θ =⇒ sin θ =x

a=⇒ θ = Sin−1

(x

a

)

. (181)

Therefore,∫

1√a2 − x2

dx = Sin−1(x

a

)

+ C . (182)

Example 2 revisited: Suppose that you remembered that

∫

1√a2 − x2

dx = Sin−1(x) . (183)

You could use this fact to find the antiderivative,

∫

1√a2 − x2

dx , (184)

as follows. First, factor out the a2 from the square root:

∫

1√a2 − x2

dx =

∫

1

a

1√

1−(

xa

)2dx . (185)

Now make the change of variable,

u =x

a=⇒ du =

1

adx =⇒ dx = adu . (186)

53

The integral at the right then becomes

1

a

∫

1√

1−(

xa

)2dx =

1

a

∫

1

1− u2a du

=

∫

1

1− u2du

= Sin−1u+ C

= Sin−1(x

a

)

+ C , (187)

in agreement with our earlier result.

Example 2: It’s possible that we also have powers of x present in the integral, e.g.,

∫

x√a2 − x2

dx . (188)

This integral is actually quite easy since the top is a derivative of the inside of the bottom. Let

u = a2 − x2 =⇒ du = −2x dx =⇒ x dx = −1

2du , (189)

so that the above integral becomes

−1

2

∫

1

u1/2du = −1

2(2)u1/2 + C

= −√

a2 − x2 +C . (190)

Example 4: The antiderivative,∫

x2√a2 − x2

dx . (191)

As before, let

x = a sin θ =⇒ dx = a cos θ . (192)

54

The above integral becomes∫

x2√a2 − x2

dx =

∫

a2 sin2 θ

a cos θa cos θ dθ

= a2∫

sin2 θ dθ

= a2∫[

1

2− cos 2θ

]

dθ (193)

=a2

2θ − a2

4sin(2θ)

=a2

2θ − a2

2sin θ cos θ

=a2

2Sin−1

(x

a

)

− a2

2

(x

a

)

√

1− x2

a2+ C

=a2

2Sin−1

(x

a

)

− x

2

√

a2 − x2 + C .

(194)

Integrals which contain√a2 + x

2

Example 5: The antiderivative,∫

√

a2 + x2 dx . (195)

With an eye to the trigonometric relation,

tan2 x+ 1 = sec2 x , (196)

let us consider the change of variable,

x = a tan θ =⇒ dx = a sec2 θ dθ . (197)

With this change of variable, the square root term becomes

√

a2 + x2 =√

a2 + a2 tan2 θ

= a√

1 + tan2 θ

= a sec θ . (198)

As earlier, the sec θ term should technically be | sec θ|, but we’ll restrict the range of θ once again to

[−π/2, π/2].

And while we’re at it, let’s draw the triangle that is associated with this change of variable, starting

with the fact that

tan θ =x

a. (199)

55

θ

x

a

√

a2 + x2

We now perform the above change of variable in the integral,

∫

√

a2 + x2 dx =

∫

(a sec θ)a sec2 θ dθ

= a2∫

sec3 θ . (200)

We evaluated this antiderivative in the previous lecture and simply employ the result here:

∫

√

a2 + x2 dx = a2∫

sec3 θ

=a2

2[sec θ tan θ + ln | sec θ + tan θ|] + C . (201)

We must now “translate” the functions in θ into functions involving x. From the diagram,

tan θ =x

asec θ =

√x2 + a2

a. (202)

Inserting these results into our expression, we obtain

∫

√

a2 + x2 dx =a2

2[sec θ tan θ + ln | sec θ + tan θ|] + C

=x

2

√

a2 + x2 +a2

2ln

∣

∣

∣

∣

∣

√a2 + x2

a+

x

a

∣

∣

∣

∣

∣

+ C . (203)

Note that we can write the logarithmic term as follows,

ln

∣

∣

∣

∣

∣

√a2 + x2

a+

x

a

∣

∣

∣

∣

∣

= ln

∣

∣

∣

∣

∣

√a2 + x2 + x

a

∣

∣

∣

∣

∣

= ln∣

∣

∣

√

a2 + x2 + x∣

∣

∣− ln |a| . (204)

The constant term ln |a| can be absorbed into the arbitrary constant C in the previous expression. As

such, our final result is

∫

√

a2 + x2 dx =x

2

√

a2 + x2 +a2

2ln∣

∣

∣

√

a2 + x2 + x∣

∣

∣+ C .

(205)

The result can be checked by differentiation. (But we won’t do it here!)

56

Example 6: The antiderivative∫

1√a2 + x2

dx . (206)

Once again, we consider the following change of variable,

x = a tan θ =⇒ dx = a sec2 θ dθ . (207)

The above integral becomes

∫

1√a2 + x2

dx =

∫

1

a sec θ(a sec2 θ) dθ

=

∫

sec θ dθ

= ln | sec θ + tan θ|+ C

= ln

∣

∣

∣

∣

∣

√a2 + x2

a+

x

a

∣

∣

∣

∣

∣

+ C

= ln |√

a2 + x2 + x|+ C , (208)

where we have once again absorbed the ln |a| term into the arbitrary constant.

Integrals which contain√x2 − a

2

Example 7: The antiderivative,∫

√

x2 − a2 dx . (209)

Once again with an eye to the trigonometric relation,

tan2 x+ 1 = sec2 x =⇒ sec2 x− 1 = tan2 x , (210)

we consider the following change of variable,

x = a sec θ =⇒ dx = a sec θ tan θ dθ . (211)

This implies thatx

a= sec θ =⇒ θ = Tan−1

(x

a

)

. (212)

Let’s draw the triangle associated with this change of variable:

57

θ

a

x √

x2− a2

Performing the change of variable in the above integral yields,

∫

√

x2 − a2 dx =

∫

(a tan θ)(a sec θ tan θ), dθ

= a2∫

tan2 θ sec θ dθ

= a2∫

(sec2 θ − 1) sec θ dθ

= a2∫

sec3 θ dθ − a2∫

sec θ dθ

= a2[

1

2sec θ tan θ +

1

2ln | sec θ + tan θ| − ln | sec θ + tan θ|

]

+ C

=a2

2[sec θ tan θ − ln | sec θ + tan θ| ] + C

=a2

2

[

(x

a

)

(√x2 − a2

a

)

− ln

∣

∣

∣

∣

∣

x

a+

√x2 − a2

a

∣

∣

∣

∣

∣

]

+ C

=x

2

√

x2 − a2 − a2

2ln |x+

√

x2 − a2|+ C , (213)

where, once again, we have absorbed ln |a| into the arbitrary constant.

58