119
C H A P T E R 4
Methods of Integration I
e integration methods we have learned thus far are based on the fact that integrals are reversed
derivatives. To some extent we can increase the reach of the “reverse derivative” technique by
setting things up with algebra. e chapters on methods of integration will introduce the most
useful and broadly applicable of thousands of integration methods that have been found over
the last several centuries.
4.1 u-SUBSTITUTION
Every integral method is a reversed derivative, and u-substitution is the reverse of the chain rule.
We introduced u-substitution in Chapter 3 to permit us to solve more complex equations with
polynomial techniques. For integrals the technique is much the same, except that we need to
worry about the differential. Let’s start with an example.
Example 4.1 Find
Z
2xe
x
2
dx:
Solution:
is is not an integral for which we already have a form. Set u D x
2
and then compute
du D 2x dx. ese pieces make up all of the integral and we can now solve the problem as
follows:
Z
2xe
x
2
dx D
Z
e
x
2
.2x dx/ Set it up
D
Z
e
u
du Substitute
D e
u
C C Integrate
D e
x
2
C C At the end, reverse the substitution.
˙
120 4. METHODS OF INTEGRATION I
Recall that you can check an integral by taking a derivative, so
e
x
2
C C
0
D e
x
2
2x D 2xe
x
2
and we see the substitution worked to let us do this integral. Notice that the chain rule re-created
du for us.
Let’s do some more examples.
Example 4.2 Find
Z
cos.2x/ dx.
Solution:
We know that the integral of cosine is sine. So the only problem is the 2x. is is where
to make our u-substitution.
If we let u D 2x then du D 2dx so dx D
1
2
du.
Now we substitute and integrate:
Z
cos.2x/ dx D
Z
cos.u/
1
2
du
D
1
2
Z
cos.u/ du
D
1
2
sin.u/ CC
D
1
2
sin.2x/ C C
˙
is last example has a very general version that can let you do some types of u-substitution in
your head.
4.1. u-SUBSTITUTION 121
Example 4.3 Suppose that F .x/ C C D
Z
f .x/ dx. en what is
Z
f .ax C b/ dx‹
Solution:
Let u D ax Cb and so du D a dx and dx D
1
a
du. en:
Z
f .a x C b/ dx D
Z
f .u/
1
a
du
D
1
a
Z
f .u/ du
D
1
a
F .u/ CC
D
1
a
F .ax C b/ C C
˙
is is important enough to put in a Knowledge Box. What we are doing is correcting for
composing a function with a linear function.
Knowledge Box 4.1
u-substitution to correct a linear composition
If
Z
f .x/ dx D F .x/ CC , then
Z
f .ax C b/ dx D
1
a
F .ax C b/ C C
Example 4.4 Find
Z
e
0:5x1
dx.
Solution:
2e
0:5x1
C C
by directly applying the linear correction with a D 0:5 and b D 1.
˙
122 4. METHODS OF INTEGRATION I
is example shows how much calculation the linear correction technique saves.
Sometimes a u substitution is not obvious.
Example 4.5 Find
Z
dx
x ln.x/
Solution:
e key to this one is remembering that
d
dx
ln.x/ D
1
x
making u D ln.x/ a natural choice to
try. en du D
dx
x
and we can integrate.
Z
dx
x
ln
.x/
D
Z
1
ln
.x/
dx
x
D
Z
1
u
du
D ln.u/ CC
D ln.ln.x// C C
˙
Sometimes the substitution is so unobvious that people clearly figured out the integral another
way. e following example has this property.
Example 4.6 Find
Z
sec.x/ dx:
Solution:
We begin with an algebraic transformation.
Z
sec.x/ dx D
Z
sec.x/
sec.x/ C tan.x/
sec.x/ C tan.x/
dx
D
Z
sec
2
.x/ C sec.x/ tan.x/
tan.x/ C sec.x/
dx
4.1. u-SUBSTITUTION 123
At this point let u D tan.x/ C sec.x/ so that
du D
sec
2
.x/ C sec.x/ tan.x/
dx
which makes this all:
Z
sec.x/ dx D
Z
du
u
D ln.juj/ CC
D ln.jtan.x/ C sec.x/j/ C C
e absolute value bars are needed because the value of these functions can be negative.
˙
We also now know enough to capture the integral of another of the basic functions.
Example 4.7 Compute
Z
tan.x/ dx.
Solution:
Since tan.x/ D
sin.x/
cos.x/
we let u D cos.x/ obtaining du D sin.x/dx or du D sin.x/dx.
Plugging in the substitution we get
Z
tan.x/ dx D
Z
du
u
D
Z
du
u
D ln.juj/ C C
D ln
1
juj
C C
D ln
1
jcos.x/j
C C
D ln.jsec.x/j/ C C
Taking the minus sign inside the log is a negative first power or reciprocal, which is how we get
to the secant.
˙
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