Session 12: Integration by Substitution
Owen Biesel
Integration by substitution is the analogue for antiderivatives of the chain rule for
derivatives. If you’re ever unsure whether you’ve done integration by substitution correctly, you can use the chain rule to check your work.
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Simple Substitutions
The simplest sorts of substitution is where you find an antiderivative for f (x + d) by
finding an antiderivative for f and evaluating it at x + d:
If b is any number and
Z
f (x) dx = F (x) + C, then
Z
f (x + b) dx = F (x + b) + C.
R
R
For example, because cos(x) dx = sin(x) + C, we can deduce that cos(x − 5) dx =
sin(x − 5) + C. This is something we can check with the chain rule:
d d ′
sin(x − 5) = sin (x − 5) ·
x−5
dx
dx
= cos(x − 5) · 1
= cos(x − 5).
Or if we want to find an antiderivative for (x + 5)2 , we can use the fact that
1 3
x + C to say
3
Z
1
(x + 5)2 dx = (x + 5)3 + C
3
1
125
= x3 + 5x2 + 25x +
+ C.
3
3
R
x2 dx =
This agrees with what we would have gotten if we had expanded (x + 2)3 and integrated
each term separately:
Z
Z
2
(x + 5) dx = x2 + 10x + 25 dx
1
= x3 + 5x2 + 25x + C.
3
The only difference is the 125/3 term, but that is okay: different antiderivative-finding
methods may produce different antiderivatives, but any two antiderivatives for the same
expression should differ only by the addition or subtraction of a constant.
1
The next simplest case is where we wish to evaluate an integral of the form
Z
f (ax + b) dx.
Then the rule looks like this:
If a and b are numbers and
Z
Z
f (x) dx = F (x) + C, then
1
f (ax + b) dx = F (ax + b) + C.
a
Once again, we can check this using the chain rule: if F ′(x) = f (x), then
1
d 1
d ax + b
F (ax + b) = F ′ (ax + b) ·
dx a
a
dx
1
= f (ax + b) · a
a
= f (ax + b).
R
For example, because ex dx = ex + C, we have
Z
1
e3x dx = e3x + C.
3
R
R
Or if we were trying to calculate (2x−3)3 dx, we could use the fact that x3 dx = 14 x4 +C
to deduce that
Z
1 1
(2x − 3)3 dx = · (2x − 3)4 + C.
2 4
2
Complicated Substitutions
The general rule for integration by substitution is the following:
If F is an antiderivative for f , and y is a variable depending on x, then
Z
dy
f (y) ·
dx = F (y) + C.
dx
For example, we could let f = cos, F = sin, and y = x2 (so that dy/dx = 2x). Then
integration by substitution tells us that
Z
cos(x2 ) · (2x) dx = sin(x2 ) + C.
Again, you can check this with the chain rule:
d d 2
sin(x2 ) = cos(x2 ) ·
x = cos(x2 ) · (2x).
dx
dx
2
R
Since we can express the fact that F is an antiderivative for f in the form f (y) dy =
F (y) + C, integration by substitution is often expressed in the following form, where it
appears that you are “canceling” the dx:
Z
Z
dy
dx = f (y) dy.
f (y) ·
dx
The game of integration by substitution is to try to find a way to write what you
want to integrate as fR(y) √
· (dy/dx) for some new variable y. For example, if we wish to
evaluate the integral 2x x2 + 5 dx we could let y = x2 + 5. Then dy/dx = 2x, so we
could write
Z p
Z
√
2
(x2 + 5) · (2x) dx
2x x + 5 dx =
Z
√ dy
dx
=
y·
dx
Z
√
=
y dy
√
Now y has a known antiderivative, since it can be expressed as the power y 1/2 . So we
can continue:
Z
1 3/2
y +C
y 1/2 dy =
3/2
2
= y 3/2 + C
3
2
= (x2 + 5)3/2 + C.
3
You can check with the chain rule that
√
2x x2 + 5:
2
(x2
3
+ 5)3/2 is indeed an antiderivative for
2
d 2
d 2 2
x +5
(x + 5)3/2 = · (3/2)(x2 + 5)1/2 ·
dx 3
3
dx
2
1/2
= 1 · (x + 5) · 2x
√
= 2x x2 + 5.
Beware, though, that it’s often not obvious what a helpful substitution will be. You
may have
to try several possible definitions of y before you can write your integral in the
R
dy
dx where f has a known antiderivative, and it may not even be possible to
form f (y) dx
do so.
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3
Review Questions
1. Which of these statements about indefinite integrals are correct? Check each by
differentiating using the chain rule.
Z
Z
1
1
5
4
x4 dx = x5 + C.”
(a) “ (cos(x)) dx = (cos(x)) + C because
5
5
Z
Z
1
1
(b) “
dx = ln(x − 3) + C because
dx = ln(x) + C.”
x−3
x
Z
Z
2
2
(c) “ cos(x ) dx = sin(x ) + C because
cos(x) dx = sin(x) + C.”
Z
Z
3x
3x
(d) “ e dx = e + C because
ex dx = ex + C.”
2. Calculate the following indefinite integrals. (For extra practice, check each of your
answers by differentiating.)
Z
(a) (5x + 2)2 dx =??
Z
π
(b)
cos(x − ) dx =??
6
Z
(c)
e2x+3 dx =??
3. Use integration by substitution to calculate the following integrals:
Z
(a) (sin(x))4 cos(x) dx. (Hint: let y = sin(x).)
Z
2
(b)
−2xe−x dx. (Hint: let y = −x2 .)
Z
3
√
dx. (Hint: let y = 3x − 8.)
(c)
3x − 8
Z
2x
(d)
dx. (Hint: let y = x2 + 1.)
2
x +1
4
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Answers
1. Which of statements about indefinite integrals are correct? Check each by differentiating using the chain rule.
R
R
(a) “ (cos(x))4 dx = 15 (cos(x))5 + C because x4 dx = 51 x5 + C.”
R
R 1
dx = ln(x − 3) + C because x1 dx = ln(x) + C.”
(b) “ x−3
R
R
(c) “ cos(x2 ) dx = sin(x2 ) + C because cos(x) dx = sin(x) + C.”
R
R
(d) “ e3x dx = e3x + C because ex dx = ex + C.”
We can calculate the derivative of each alleged antiderivative using the chain rule:
4
4
R
d
1
d
(a) dx
(cos(x))5 = 51 ·5(cos(x))4 · dx
(cos(x)) = cos(x) − sin(x) , so cos(x) dx
5
5
5
4
cannot be 51 cos(x) +C, because the derivative of 15 cos(x) is not cos(x) .
d
dx
d
1
1
· dx
(x − 3) = x−3
· 1 = x−3
, so ln(x − 3) is indeed an
R 1
and x−3 dx = ln(x − 3) + C.
R
d
d
(c) dx
(sin(x2 )) = cos(x2 )· dx
(x2 ) = cos(x2 )·2x, so cos(x2 ) dx cannot be sin(x2 )+
C, because the derivative of sin(x2 ) is not cos(x2 ).
R
d
d
(e3x ) = e3x · dx
(3x) = e3x · 3, so e3x dx cannot be e3x + C because the
(d) dx
derivative of e3x is not e3x .
(b)
(ln(x − 3)) =
antiderivative of
1
x−3
1
x−3
So only the second indefinite integral, where x is replaced by x − 3, is correct; all
the other indefinite integrals have replaced x by something whose derivative with
respect to x is not 1.
2. Calculate the following indefinite integrals. (For extra practice, check each of your
answers by differentiating.)
R
(a) (5x + 2)2 dx =??
R
(b) cos(x − π6 ) dx =??
R
(c) e2x+3 dx =??
R
For each case, we use the fact that f (ax + b) = a1 F (ax + b) + C if F is an
antiderivative of f :
R
(a) Since an antiderivative for x2 is 13 x3 , we have (5x+2)2 dx = 15 · 31 (5x+2)3 +C.
R
R
(b) Similarly, since cos(x) dx = sin(x) + C, we have cos(x − π6 ) dx = sin(x −
d
π
) + C. No extra factor is necessary, since dx
(x − π6 ) = 1.
6
R x
R
(c) Since e dx = ex + C, we have e2x+3 dx = 21 e2x+3 + C.
3. Use integration by substitution to calculate the following integrals:
R
(a) (sin(x))4 cos(x) dx. (Hint: let y = sin(x).)
R
2
(b) −2xe−x dx. (Hint: let y = −x2 .)
R 3
dx. (Hint: let y = 3x − 8.)
(c) √3x−8
R 2x
(d) x2 +1 dx. (Hint: let y = x2 + 1.)
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(a) If we let y = sin(x) and dy/dx = cos(x), then
Z
Z
Z
1
1
4
4 dy
(sin(x)) cos(x) dx = y
dx = y 4 dy = y 5 + C = (sin(x))5 + C.
dx
5
5
(b) If we let y = −x2 , then dy/dx = −2x. So
Z
Z
Z
dy y
2
−x2
e dx = ey dy = ey + C = e−x + C.
−2xe
dx =
dx
(c) If we let y = 3x − 8, then dy/dx = 3. Then
Z
Z
3
dy/dx
√
dx =
√ dx
y
3x − 8
Z
1
=
√ dy
y
Z
= y −1/2 dy
1 +1/2
y
)+C
+1/2
= 2(3x − 8)1/2 + C
=(
(d) Letting y = x2 + 1, we have dy/dx = 2x. Then
Z
Z
Z
dy/dx
1
2x
dx =
dx =
dy = ln |y| + C = ln |x2 + 1| + C.
2
x +1
y
y
In fact, since x2 + 1 is always positive, we can write this antiderivative as
ln(x2 + 1), without the absolute value bars.
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