CHAPTER 4
SECTION 4.5
INTEGRATION BY
SUBSTITUTION
Theorem 4.12 Antidifferentiation of
a Composite Function
Substitution with Indefinite Integration
• This is the “backwards” version of the
chain rule
• Recall …

d 2
x  4x  7
dx

5

 5 x  4x  7
2
 2x  4 
4
• Then …
 5 x
2
 4x  7
  2 x  4  dx   x
4
2

5
 4x  7  C
Substitution with Indefinite Integration

f ( x)dx
• In general we look at the f(x) and “split” it
– into a g(u) and a du/dx
du
f ( x)  g (u )
dx
• So that …
du
g
(
u
)
dx

G
(
u
)

C

dx
Substitution with Indefinite Integration
• Note the parts of the integral from our
example
 5 x
2
 4x  7
  2 x  4  dx   x
4
2

5
 4x  7  C
du
g
(
u
)
dx

G
(
u
)

C

dx
Substitution with Indefinite Integration
• Let u = 
 5 x
2
x2  4 x  7
 So, du = (2x -4)dx
 4x  7
  2 x  4  dx   x
4
2
5u
du

u

C

4
5

5
 4x  7  C
Guidelines for Making a Change
of Variables
Theorem 4.13 The General Power
Rule for Integration
Example 1:
  x  2
5
dx
 u du
Let u  x  2
du  dx
5
1 6
u C
6
 x  2
6
6
C
The variable of integration
must match the variable in
the expression.
Don’t forget to substitute the value
for u back into the problem!

Example 2:

1  x  2 x dx
2
One of the clues that we look for is
if we can find a function and its
derivative in the integrand.
1 x
2
Let u  1  x
2
The derivative of
u
1
2
du
2
u C
3

2
1 x
3

2x dx .
du  2x dx
3
2
3
2 2
is
C
Note that this only worked because
of the 2x in the original.
Many integrals can not be done by
substitution.

Example 3:

4 x  1 dx
Let u  4x 1
du  4 dx
1
2
1
 u  4 du
3
2
1
du  dx
4
Solve for dx.
2
1
u  C
3
4
3
2
1
u C
6
3
1
 4 x  1 2  C
6

Example 4:
 cos  7 x  5 dx
1
 cos u  7 du
Let u  7 x  5
du  7 dx
1
du  dx
7
1
sin u  C
7
1
sin  7 x  5   C
7

Example 5:
 
2
3
x
sin
x
dx

1
sin u du

3
1
 cos u  C
3
Let u  x3
du  3x 2 dx
1
2
du  x dx
3
2
We solve for x dx
because we can find it
in the integrand.
1
3
 cos x  C
3

Example 6:
4
sin
 x  cos x dx
 sin x 
u
4
4
cos x dx
du
Let u  sin x
du  cos x dx
1 5
u C
5
1 5
sin x  C
5

Can You Tell?
• Which one needs substitution for
integration?
x
3
x

5
dx



2
x
3
x

5
dx



2
5
Integration by Substitution
1.
x
2

5
 1 dx
Integration by Substitution
1.
 x x
2

5
 1 dx 
1
2
x
2

5
 1 2 xdx

u  x 1

du
 2x
dx

2
du  2xdx
1
2
1
12
1
12
 u 
du
u 
C
5
x
2
6

6
1 C
2.
 5 cos  5x  dx
3.
 x x
2
3

4
 1 dx
2.
 5 cos  5x  dx
u  5x
du
5
dx
du  5dx
3.
x
2


  cos  5 x  5dx
  cos  u  du
 sinu  C
 sin  5x   C


4
2
x  1 dx   x  1 x dx
3
4
u  x 1
du
 3x 2
dx
3
du  3 x 2dx
3


 u  du

1
3

1
3

1
15
4
x  1 3 x 2dx
3
4
u 
5
C 
1
15
x
3

5
1 C
4.
2
sin
  3 x  cos  3 x  dx
4.
 sin  3 x  cos  3 x  dx   sin  3 x  3cos 3 x  dx
u  sin  3x 
  u du
du
2
 3 cos  3 x 
dx
du  3cos  3x  dx
2
1
3
1
3
2
 91 u 3  C

1
9
 sin 3x 
3
C
Solve the differential equation
dy
10 x 2

dx
1 x3
Solve the differential equation
dy
10 x 2

dx
1 x3
dy 
10 x 2
1 x 3
u  1 x
du
 3x 2
dx
du  3 x 2dx
3
dx
x2

dy  10

10
2
dy 
3
x
dx

3
3
1 x
10  12
dy 
u du

3

y
y
1 x 3
1
dx
10
3
 2u  C
20
3
1 x 
1
2
3
1
2
C
Theorem 4.14 Change of Variables
for Definite Integrals
2
1.
 x x
0
2

2
 1 dx
2
1.
 x x
0

2
2
 1 dx 
2
u  x 1
du
 2x
dx
du  2xdx
1
2
x
0
x 2
2

1
2
u  0 1 1
u  22  1  5
3 5
 61  u  

1

2
 1 2xdx
 u 
2
du
x 0
x 2


3 2
2

1
 61  u  

x

1

 x 0 6 
0
 61 125  1
or you could convert the bound to u’s.
2
2
3

1
6
124
62

3
Example 7:


4
0
tan x sec2 x dx
new limit
1
 u du
0
new limit
1
1 2
 u
2 0
When
The technique is a little different
for definite integrals.
We can
find new
Let u  tan x
limits, and
then we
du  sec 2 x dx
don’t have
to
x = 0, u = tan 0  0, substitute
back.
When x =

4
, u  tan

4
1
1

2

Example 9:

1
1
3x
Let u  x3  1
x  1 dx
2
3
du  3x dx
2

2
0
1
2
u  1  0
u 1  2
u du
2
u
3
3 2
2
2
2
3
Don’t forget to use the new limits.
0
3
2
2
 2 2
3
4 2

3

Theorem 4.15 Integration of Even and
Odd Functions
Even/Odd Functions
If f(x) is an even function, then
a
 f  x  dx 
a
If f(x) is an odd function, then
a
 f  x  dx 
a
2
3.

2

x 3  x 2  3 x  2 dx
Even/Odd Functions
If f(x) is an even function, then
a
a
a
0
 f  x  dx  2 f  x  dx
If f(x) is an odd function, then
a
 f  x  dx 
0
a
2
3.
 x
2
3

 x  3 x  2 dx
2
2

2
2
2
2
2
  x  dx    x  dx    3x  dx    2 dx
3
2
2
2
2
2
 
 0  2 x 2 dx  0  2  2  dx
0
 2  31 x  2x 
3
0
2
0
 2  83  4   0

40
3