x = (A + B)x 2 + (2B + C)x + (A + 2C).

This equation must be identically true, so we ¬nd (identifying powers of x) that

A+ B =0

2B + C = 1

+ 2C = 0

A

Solving this system, we ¬nd that A = ’2/5, B = 2/5, C = 1/5. So

’2/5 (2/5)x + (1/5)

x dx

= dx + dx

x+2

x 3 + 2x 2 + x + 2 x2 + 1

’2 1 2x 1 dx

= log |x + 2| + dx + dx

x2 + 1 x2 + 1

5 5 5

’2 1 1

= log |x + 2| + log |x 2 + 1| + arctan x + C.

5 5 5

You Try It: Calculate the integral

1 dx

.

x 3 + 6x 2 + 9x

0

You Try It: Calculate the integral

dx

.

x3 + x

7.3 Substitution

Sometimes it is convenient to transform a given integral into another one by means

of a change of variable. This method is often called “the method of change of

variable” or “u-substitution.”

208 CHAPTER 7 Methods of Integration

To see a model situation, imagine an integral

b

f (x) dx.

a

If the techniques that we know will not suf¬ce to evaluate the integral, then we might

attempt to transform this to another integral by a change of variable x = •(t). This

entails dx = • (t) dt. Also

x = a ←’ t = • ’1 (a) x = b ←’ t = • ’1 (b).

and

Thus the original integral is transformed to

• ’1 (b)

f (•(t)) · • (t) dt.

• ’1 (a)

It turns out that, with a little notation, we can make this process both convenient

and straightforward.

Y

We now illustrate this new paradigm with some examples. We begin with an

inde¬nite integral.

FL

EXAMPLE 7.8

AM

Evaluate

[sin x]5 · cos x dx.

TE

SOLUTION

On looking at the integral, we see that the expression cos x is the derivative

of sin x. This observation suggests the substitution sin x = u. Thus cos x dx =

du. We must now substitute these expressions into the integral, replacing all

x-expressions with u-expressions. When we are through with this process, no

x expressions can remain. The result is

u5 du.

This is of course an easy integral for us. So we have

u6

[sin x] · cos x dx = u du = + C.

5 5

6

Now the important ¬nal step is to resubstitute the x-expressions in place of

the u-expressions. The result is then

sin6 x

[sin x] · cos x dx = + C.

5

6

209

CHAPTER 7 Methods of Integration

Math Note: Always be sure to check your work. You can differentiate the answer

in the last example to recover the integrand, con¬rming that the integration has

been performed correctly.

EXAMPLE 7.9

Evaluate the integral

3

2x x 2 + 1 dx.

0

SOLUTION

We recognize that the expression 2x is the derivative of x 2 +1. This suggests

the substitution u = x 2 + 1. Thus du = 2x dx. Also x = 0 ←’ u = 1 and

x = 3 ←’ u = 10. The integral is thus transformed to

10 √

u du.

1

This new integral is a bit easier to understand if we write the square root as

a fractional power:

10

2 · 103/2 2

10 u3/2 103/2 13/2

du = = ’ = ’.

1/2

u

3/2 3/2 3/2 3 3

1 1

You Try It: Evaluate the integral

5 dx

.

x · log |x|

3

Math Note: Just as with integration by parts, we always have the option of ¬rst

evaluating the inde¬nite integral and then evaluating the limits at the very end. The

next example illustrates this idea.

EXAMPLE 7.10

Evaluate

π/2

cos x

dx.

sin x

π/3

SOLUTION

Since cos x is the derivative of sin x, it is natural to attempt the substitution

u = sin x. Then du = cos x dx. [Explain why it would be a bad idea to let

u = cos x.] We ¬rst treat the improper integral. We ¬nd that

cos x du

dx = = log |u| + C.

sin x u

210 CHAPTER 7 Methods of Integration

Now we resubstitute the x-expressions to obtain

cos x

dx = log | sin x| + C.

sin x

Finally we can evaluate the original de¬nite integral:

π/2

π/2 cos x

dx = log | sin x|

sin x

π/3 π/3

√

3

= log | sin π/2| ’ log | sin π/3| = log 1 ’ log

2

1

= ’ log 3 + log 2.

2

You Try It: Calculate the integral

3 t dt

.

(t 2 + 1) log(t 2 + 1)

’2

7.4 Integrals of Trigonometric Expressions

Trigonometric expressions arise frequently in our work, especially as a result of

substitutions. In this section we develop a few examples of trigonometric integrals.

The following trigonometric identities will be particularly useful for us.

I We have

1 ’ cos 2x

sin2 x = .

2

The reason is that

cos 2x = cos2 x ’ sin2 x = 1 ’ sin2 x ’ sin2 x = 1 ’ 2 sin2 x.

II We have

1 + cos 2x

cos2 x = .

2

The reason is that

cos 2x = cos2 x ’ sin2 x = cos2 x ’ 1 ’ cos2 x = 2 cos2 x ’ 1.

Now we can turn to some examples.

211

CHAPTER 7 Methods of Integration

EXAMPLE 7.11

Calculate the integral

cos2 x dx.

SOLUTION

Of course we will use formula II. We write

1 + cos 2x

cos2 x dx = dx

2

1 1

= dx + cos 2x dx

2 2

1

x

= + sin 2x + C.

24

EXAMPLE 7.12

Calculate the integral

sin3 x cos2 x dx.

SOLUTION

When sines and cosines occur together, we always focus on the odd power

(when one occurs). We write

sin3 x cos2 x = sin x sin2 x cos2 x = sin x 1 ’ cos2 x cos2 x

= cos2 x ’ cos4 x sin x.

Then

sin3 x cos2 dx = cos2 x ’ cos4 x sin x dx.

A u-substitution is suggested: We let u = cos x, du = ’ sin x dx. Then the

integral becomes

u3 u5

’ u ’ u du = ’ + + C.

2 4

3 5

Resubstituting for the u variable, we obtain the ¬nal solution of

cos3 x cos5 x

sin x cos dx = ’ + + C.

3 2

3 5

You Try It: Calculate the integral

sin2 3x cos5 3x dx.

212 CHAPTER 7 Methods of Integration

EXAMPLE 7.13

Calculate

π/2

sin4 x cos4 x dx.

0

SOLUTION

Substituting

1 ’ cos 2x 1 + cos 2x

sin2 x = cos2 x =

and

2 2

into the integrand yields

2 2

1 ’ cos 2x 1 + cos 2x

π/2

· dx

2 2

0

π/2

1

= 1 ’ 2 cos2 2x + cos4 2x dx.

16 0

Again using formula II, we ¬nd that our integral becomes

1 + cos 4x 2

π/2

1

1 ’ [1 + cos 4x] + dx

16 2

0

π/2

1 1

= 1 ’ [1 + cos 4x] + 1 + 2 cos 4x + cos2 4x dx.

16 0 4

Applying formula II one last time yields

1 + cos 8x

π/2

1 1

1 ’ [1 + cos 4x] + 1 + 2 cos 4x + dx

16 4 2

0

π/2

1 1 1 1 sin 8x

x

= ’ sin 4x + x + sin 4x + +

16 4 4 2 2 16 0

1 1π 1

π

= ’0 + +0+ +0 ’ ’0 + (0 + 0 + 0 + 0)

16 42 4 4

3π

= .

256

You Try It: Calculate the integral

π/3

sin3 s cos3 s ds.

π/4

You Try It: Calculate the integral

π/3

sin2 s cos4 s ds.

π/4

213

CHAPTER 7 Methods of Integration

Integrals involving the other trigonometric functions can also be handled with

suitable trigonometric identities. We illustrate the idea with some examples that are

handled with the identity

sin2 x + cos2 x

sin2 x 1

tan x + 1 = +1= = = sec2 x.

2

2x 2x 2x

cos cos cos

EXAMPLE 7.14

Calculate

tan3 x sec3 x dx.

SOLUTION

Using the same philosophy about odd exponents as we did with sines and

cosines, we substitute sec2 x ’ 1 for tan2 x. The result is

tan x sec2 x ’ 1 sec3 x dx.

We may regroup the terms in the integrand to obtain

sec4 x ’ sec2 x sec x tan x dx.

A u-substitution suggests itself: We let u = sec x and therefore du =

sec x tan x dx. Thus our integral becomes

u5 u3

u ’ u du = ’ + C.

4 2

5 3

Resubstituting the value of u gives

sec5 x sec3 x

tan x sec x dx = ’ + C.

3 3

5 3

EXAMPLE 7.15

Calculate

π/4

sec4 x dx.

0

214 CHAPTER 7 Methods of Integration

SOLUTION

We write

π/4 π/4

sec x dx = sec2 x · sec2 x dx

4

0 0

π/4

= (tan2 x + 1) sec2 x dx.

0

Letting u = tan x and du = sec2 x dx then gives the integral

1

1 u3

u + 1 du = +u

2

3

0 0

4

=.

3

You Try It: Calculate the integral

2π

sin6 x cos4 x dx.

π

Further techniques in the evaluation of trigonometric integrals will be explored

in the exercises.

Exercises

1. Use integration by parts to evaluate each of the following inde¬nite

integrals.

log2 x dx

(a)

x · e3x dx

(b)

x 2 cos x dx

(c)

(d) t sin 3t cos 3t dt

(e) cos y ln(sin y) dy

x 2 e4x dx

(f)

215

CHAPTER 7 Methods of Integration

2. Use partial fractions to evaluate each of the following inde¬nite integrals.

dx

(a)