Integration by
Substitution
The chain rule allows us to
differentiate a wide variety of
functions, but we are able to find
antiderivatives for only a limited
range of functions? We can
sometimes use substitution or
change of variable to rewrite
functions in a form that we can
integrate.

Example 1:
  x  2
5
dx
  u du
Let u  x  2
du  dx
5
1 6
 u c
6
( x  6)

c
6
6
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 integral.
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 integral.
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

Example 7:

The technique is a little different
for definite integrals.

4
0
tan x sec2 x dx
new limit

1
0
Let u  tan x
du  sec 2 x dx
u du
new limit
1
1 2 
 u 
 2 0
1

2
u  0  tan 0  0

 
u    tan  1
4
4
We can find
new limits,
and then we
don’t have
to substitute
back.
We could have substituted back and
used the original limits.

Example 7 continued:

Using the original limits:

tan x sec2 x dx
4
0

Let u  tan x

4
0
du  sec 2 x dx
u du
 u du
Leave the
limits out until
you substitute
back.
1 2
 u
2

4
1
2
  (tan x) 
2
0
Wrong!
1
 1
2
  tan    tan 0 
The
2  limits
4 don’t
2 match!
2
1 2 1 2
 1   0
2
2
1

2
This is
usually
more work
than finding
new limits

Example 8:

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
3
2
2
3
2
2
 2 2
3
2 
 u 
 3 0
2
 2
3
Don’t forget to use the new limits.
4 2

3

Acknowledgement
I wish to thank Greg Kelly from Hanford High School, Richland, USA for his
hard work in creating this PowerPoint.
http://online.math.uh.edu/
Greg has kindly given permission for this resource to be downloaded from
www.mathxtc.com and for it to be modified to suit the Western Australian
Mathematics Curriculum.
Stephen Corcoran
Head of Mathematics
St Stephen’s School – Carramar
www.ststephens.wa.edu.au
