Integration by Substitution
Lesson 5.5
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
Example
• Try this …
 what is the g(u)?
 what is the du/dx?
(4
x

5)
dx

• We have a problem …
2
Where is the 4
which we need?
Example
• We can use one of the properties of
integrals
 c  f ( x)dx  c   f ( x)dx
Where did the 1/3
come from?
Why is this
inside now
anda 3?
a
• We will insert a factor of 4
factor of ¼ outside to balance the result
1
1 1
2
3



(4
x

5)
4
(4
x

5
)
dx

4 3
4
Can You Tell?
• Which one needs substitution for
integration?


x
3
x

5
dx

2


x
3
x

5
dx

2
5
• Go ahead and do the integration.
Try Another …

3t  1 dt
sin
x
cos
x
dx

3
Assignment A
• Lesson 5.5
• Page 340
• Problems:
1 – 33 EOO
49 – 77 EOO
Change of Variables
• We completely rewrite the integral in terms
of u and du
x  2 x  3 dx
• Example:

• So u = 2x + 3 and du = 2x dx
• But we have an x in the integrand
 So we solve for x in terms of u
u 3
x
2
Change of Variables
• We end up with
1
2
1
x

2
x

3
dx

u

3

u
du




2
1
2
• It remains to distribute the u and proceed
with the integration
• Do not forget to "un-substitute"
What About Definite Integrals
2
• Consider a variation
3t  1 dt
of integral from
previous slide
1
• One option is to change the limits

 u = 3t - 1
Then when t = 1, u = 2
when t = 2, u = 5
 Resulting integral
5
1
2
1
u du

32
What About Definite Integrals
• Also possible to "un-substitute" and use the
original limits
1
2
3
2
1
1 2
2
u
du


u

3
t

1



3
3 3
9
3
2
2
1
Integration of Even & Odd
Functions
• Recall that for an even function f ( x)  f ( x)
 The function is symmetric about the y-axis
a
• Thus

a
a
f ( x) dx  2  f ( x ) dx
0
• An odd function has f ( x)   f ( x)
 The function is symmetric about the orgin
a
• Thus

a
f ( x) dx  0
Assignment B
• Lesson 5.5
• Page 341
• Problems:
87 - 109 EOO
117 – 132 EOO