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Identifying numerators leads to
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

= .
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

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)

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