5.5 The Substitution Rule
The Substitution Rule
• If u=g(x) is differentiable function whose range is an
interval I, and f is continuous on I, then

f ( g ( x)) g '( x)dx   f (u )du
• The Substitution rule is proved using the Chain Rule for
differentiation.
• The idea behind the substitution rule is to replace a
relatively complicated integral by a simpler integral.
This is accomplished by changing from the original
variable x to a new variable u that is a function of x.
• The main challenge is to think of an appropriate
substitution.
Example
  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!
Substitution rule examples
Example
To compute
e
2x
dx set u  2 x.
Then du  2dx  dx 
1
du and
2
1 u
1 u
e
du

e C

2
2
1
 e2 x  C
2
2x
e
 dx 
We computed du by
straightforward
differentiation of the
expression for u.
The substitution u = 2x was suggested by the function to be integrated.
The main problem in integrating by substitution is to find the right
substitution which simplifies the integral so that it can be computed by the
table of basic integrals.
Example about choosing the substitution
Example
Compute

x 1
dx.
x
Here one is tempted to try the substitution u  x  1, x  u  1,
dx  du. One gets
Solution

x 1
u
dx  
du. This is not any easier.
x
u 1
Substitute u  x  1. Then x  1  u 2 ,
dx
This rewriting allows us to
and du 
 dx  2udu.
2 x 1
finish the computation using
basic formulae.

x 1
u
1 

dx  
2
udu

2
1

du  2u  2arctan(u )  C

2
2 

x
1 u
 1 u 
Next substitute back to
the original variable.
Hence

x 1
dx  2 x  1  2arctan( x  1)  C
x
The substitution rule for
definite integrals
If g’ is continuous on [a,b], and f is
continuous on the range of u=g(x) then
b
g (b )
 f ( g ( x)) g '( x)dx  
a
g (a)
f (u )du
The substitution rule for
definite integrals
e
ln x
Compute 
dx.
x
1
Example
The substitution t  ln x is suggested by
ln x
x
the function to be integrated. We have
x  1  t  0 and x  e  t  1.
1
ln x
t 
1
Hence 
dx   tdt    .
x
2 0 2
1
0
e
1
2
1
e
The area of the yellow domain is ½.
Example


4
0
tan x sec2 x dx
new limit

1
0
u du
new limit
1
1 2
u
2 0
1
2
Let u  tan x
du  sec 2 x dx
u  0  tan 0  0

 
u    tan  1
4
4
Find new
limits
Example

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
Integrals of Even and Odd Functions
Theorem
Assume that f is an even function, i.e., that
f   x   f  x  x. Then
a
a
a
0
 f  x  dx  2 f  x  dx.
Theorem
Assume that f is an odd function, i.e., that
f   x    f  x  x. Then
a
 f  x  dx  0.
a
Integrals of Even and Odd Functions

Problem
Compute
Solution

x 5 cos7 xdx.

Observe that the function x 5 cos7 x is odd.
Since the interval of integration is symmetric with

respect to the origin,

x 5 cos7 x dx  0.

An odd function is symmetric with respect
to the origin. The definite integral from -a
to a, in the case of the function shown in
this picture, is the area of the blue domain
minus the area of the red domain. By
symmetry these areas are equal, hence
the integral is 0.
-a
a