6.2
Integration by
Substitution &
Separable
Differential
Equations
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 to rewrite functions in a form that we
can integrate.

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!

Example:

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:

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:
 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:
 
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:
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:

The technique is a little different
for definite integrals.

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
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:

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

1
2 4
  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:

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
