Integration by Substitution
In this chapter we expand our methods of antidifferentiation. We have encoun
tered integrals which we have been unable to determine because we did not know
an antiderivative for the integrand. The technique we discuss below is simply
reversing the chain rule for derivatives. Remember, every derivative rule can be
reversed to create an antidifferentiation rule. Our objectives are to
• Use pattern recognition to find an indefinite integral.
• Use a change of variables to find an indefinite integral.
• Use a change of variables to evaluate a definite integral.
All are techniques for integrating composite functions. The discussion is split
into two parts: pattern recognition and change of variables. Both techniques in
volve a socalled
u
substitution. With pattern recognition we perform the substi
tution mentally, and with change of variables we write the substitution in detailed
steps.
Pattern Recognition
Recall that the
chain rule
for derivatives of composite functions states:
THEOREM
3
.
1
(Chain Rule)
.
If
y
=
F
(
u
)
and
u
=
g
(
x
)
are differentiable then
d
dx
[
F
(
g
(
x
))] =
F
0
(
g
(
x
))
g
0
(
x
)
or, equivalently,
d
dx
[
F
(
u
)] =
F
0
(
u
)
du
dx
.
Notice in this setup that
du
dx
=
g
0
(
x
)
(
3
.
1
)
the corresponding differentials are
1
1
Though
du
dx
is a derivative and a single
entity, when working with differentials
it is as if one can multiply each side
of (
3
.
1
) by
dx
to obtain the equation in
(
3
.
2
).
du
=
g
0
(
x
)
dx
.
(
3
.
2
)
Reversing the chain rule, from the definition of an antiderivative, we get:
THEOREM
3
.
2
(Substitution)
.
Assume
F
is an antiderivative of
f
(so
F
0
=
f
) and that
g
is
differentiable. Then letting
u
=
g
(
x
)
so
du
=
g
0
(
x
)
dx
, we get
Z
f
(
g
(
x
))
g
0
(
x
)
dx
=
Z
f
(
u
)
du
=
F
(
u
) +
C
=
F
(
g
(
x
)) +
c
.
(
3
.
3
)
math 131
,
days 9 and 10
u

substitution
2
Once we know
F
is and antiderivative of
f
, then by definition of antiderivative
we have
R
f
(
u
)
du
=
F
(
u
) +
c
. . . for example,
R
cos
u du
=
sin
u
+
c no matter what
the differentiable function u is
. This is the whole point of substitution whether it is
done mentally or written out: Convert the problem into one that we can solve ‘just
by looking at it.’
To apply Theorem
3
.
2
directly, we must recognize the presence of
f
(
g
(
x
))
and
g
0
(
x
)
, i.e.,
f
(
u
)
and
du
dx
. Note that the composite function in the integrand has an
outside function f
and an
inside function g
. Moreover, the derivative
g
0
(
x
)
must also
be present as a factor of the integrand so that (
3
.
3
) becomes:
Z
outter
z}{
f
(
g
(
x
)

{z
}
inner
)
inner deriv
z
}
{
g
0
(
x
)
dx
.
EXAMPLE
3
.
1
(Recognizing the pattern)
.
Determine
Z
cos
(
x
3
)
3
x
2
dx
.
SOLUTION.
There is a function which is literally ‘inside’ the parentheses in this prob
lem. So the inside function is
u
=
x
3
and its derivative, 3
x
2
is a factor in the inte
grand.
Z
cos
(
u
z}{
x
3
)
du
z
}
{
3
x
2
dx
.
In other words,
2
the integral can be thought of as
2
Let
u
=
x
3
and
du
=
3
x
2
dx
.