Integration by Substitution
‫التكامل بالتعويض‬
Pattern recognition
Chain rule:
d
F ( g ( x))  F ' ( g ( x)) g ' ( x)
dx
Antiderivative of chain rule:
 F ' ( g ( x)) g ' ( x)dx  F ( g ( x)) C
In other words….
Integral of f(u)du where u is the inside function
and du is the derivative of the inside function…
If u = g(x), then du = g’(x)dx and

f (u )du  F (u ) C
Substitution Rule
d
f  g  x    f   g  x   g  x 
dx
 f   g  x   g  x  dx  f  g  x    C
 f  g  x  g  x  dx   f u  du, where u  g  x , du  g  x  dx
Chain Rule
If y = f(u) is a differentiable function of u and u = g(x) is a differentiable
function of x, then y = f(g(x)) is a differentiable function of x and
dy
dy du


dx
du dx
or, equivalently,
d
f ( g ( x ))  f '( g ( x )) g '( x )

dx
Think of the composite function as having 2 parts – an inner part and an
outer part.
outer
y  f ( g ( x ))  f ( u)
inner
Example:
(Exploration 1 in the book)

1  x  2 x dx
2
is
u
1
2
2
u C
3


2x dx
2
du  2x dx
3
2
2
1 x
3
1 x
2
Let u  1  x
du
3
2 2
.
C

Example
2
4
x(x
3)
dx

2
u  x  3 so,

du
 2x
dx
du
and dx 
2x
1
4 du
xu
u4 du
 2x

2

5
5
u

10


x

c 
2
 3
10
c
Example: (Not in book)
 
2
3
x
sin
x
dx

1
sin u du

3
Let u  x3
du  3x 2 dx
1
2
du  x dx
3
1
 cos u  C
3
1
3
 cos x  C
3

Example 7:
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

Compute

x 5 cos7 xdx.

x 5 cos7 x is odd.

,


x 5 cos7 x dx  0.
-a
a
Model Problem
Evaluate
A =
 2
 2
 sin
3

cos x  sin x cos x dx= 0
1
0.5
-1
1
-0.5
-1
f ( x )  sin 3 x cos x  sin x cos x
f (  x )  sin 3   x  cos   x   sin   x  cos   x 
f (  x )   sin 3 x cos x  sin x cos x   f ( x )
Example 8:


4
0
tan x sec2 x dx
new limit

1
0
u du
new limit
1
1 2
u
2 0
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.
1
2

Example
e
ln x
1 x dx.
t  ln x
ln x
x
x  1  t  0 and x  e  t  1.
1
ln x
t 
1
1 x dx  0 tdt  2   2 .
0
e
1
2
1
e
Example

x
x  3
2
dx
du
u  x  3,
1,dx  du
dx


u x3
x  u 3
x
du
2
u

u 3
du 
2
u
1 3 
 u  u2 du
3
 ln u   c
u
3
 ln x  3 
c
x  3

Model Problem
Evaluate A = 
5
1
u
x
dx
2x  1
u2  2 x  1
2x  1
u2  1
x
2
dx  u du
determine new upper and lower limits of integration
lower limit
upper limit
When x = 1, u = 1
x = 5, u = 3


3
1
1  u2  1 
u du


u
2 
3
1 3 2
  u  1 du
2 1
1
1

 16
1 u

9

3


1

 
 u


2
3
3
2 3

1
3
Model Problem
Before substitution
After substitution
7
5
6
4
5
3
4
3
2
2
1
1
2
4
6
2
A =
5
1
x
dx
2x  1
Area of region is 16/3
=
A =
3
1
4
u2  1
du
2
Area of region is 16/3
Example
x

2



  x  2
u  1 x
1  x dx 
du  1 dx
1  x  dx  
  3  u  u
   3u
1/ 2
2u
1/ 2
1 u  x
dx 
u
3/ 2
u 1  x
3/ 2
 u
2
5
2 1  x 
3u  x  2
 dx 
5/ 2
C
3/ 2

2
5
1  x 
5/ 2
C
Integration by Substitution
sin
x
cos
x
dx


3
Integration by Substitution
sec
x
tan
x
dx


2
Integration by Substitution
3
x
5
x

4
dx




7
8
4
Alternative approach …
Example 2

Let u  7 x  5

3
3
7 x  5 dx  ?

du  7dx

1
du  dx
7
1
1 13
7 x  5 dx   u  du   u du
7
7
13
1 3 43
  u c
7 4
3
43
  7 x  5  c
28