Aim: What is Integration by Substitution?
Do Now:

Find the derivative of f ( x )  3 x  2 x
Aim: Integration by Substitution
Course: Calculus
2

3
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
Aim: Integration by Substitution
Course: Calculus
Do Now

Find the derivative of f ( x )  3 x  2 x
2
Then f ( x )   u 
Let u = 3x –
n 1 du
dy
General Power Rule
 n  u  x  
dx
dx
3
2x2
u’
un - 1
n

2


2
 3  4x
f '  3 3x  2x
f '  3 3x  2x
Aim: Integration by Substitution
2
d
 3 x  2 x 2 
dx
2
Course: Calculus

3
u - substitution

dy
2
Find
for y  x  1
dx

3
d
f ( g( x ))  f '( g( x )) g '( x )

dx
y  f ( g( x ))
u  g( x )
y  f ( u)

y  x 1
2

3
yu
u  x 1
2
dy dy du


dx du dx
 3u 2  2 x

chain rule

2
 3 x  1 2x
2

 6x x  1
2
3

Aim: Integration by Substitution
2
substitute
&
simplify
 6x x
2

2
 1 dx  ?
Course: Calculus
u – substitution in Integration
From the definition of an antiderivative it
follows that:
 F '( g( x )) g '( x ) dx  F  g( x )   C
 F  u  C
Let g be a function whose range is an interval I,
and let f be a function that is continuous on I. If g
is differentiable on its domain and F is an
antiderivative of f on I, then
 f ( g( x )) g '( x ) dx  F  g( x )   C
If u  g ( x ), then du  g '( x )dx and
 f (u) du  F (u)  C
Aim: Integration by Substitution
Course: Calculus
Recognizing the Pattern
g(x) 2 g’(x)
3
1 g(x)
dy
y
x2  1  4
 x2  1 2 x
3
dx




inside function g(x)
g(x)
g’(x)
3
1 2
2
2
x 1 C
Evaluate   x  1  2 x  dx 
3


(g(x))2
Recognizing the f '( g( x )) g '( x ) Pattern
g(x) = x2 + 1
g’(x) = 2x
f ( g ( x ))  ( g( x ))2  f ( x 2  1)
Aim: Integration by Substitution
Course: Calculus
Model Problem
g’ g(x)
 cos xdx  sin x  C
Evaluate  5cos5x dx
Recognize the f '( g( x )) g '( x ) Pattern?
f ' uu '
What is the inside function, u? 5x
f ( x )  cos( g( x ))
cos(g(x)) g’
Evaluate  cos5 x 5 dx  sin5x  C
Check
Aim: Integration by Substitution
 cos axdx 
sin ax
C
a
 kf ( x )dx  k  f ( x )dx
Course: Calculus
Multiplying/Dividing by a Constant
 kf ( x ) dx  k  f ( x ) dx

g(x)
Evaluate  x x  1
2

2
dx
Recognize the f '( g( x )) g '( x ) Pattern?
What is the inside function? x2 + 1
f ( x )  ( g ( x ))2
g '( x )  2 x
problem!
factor of 2 is missing
 x x
2
1

2
k!
1
dx    x  1   2 x  dx
 2
Aim: Integration by Substitution
2
2
Course: Calculus
Multiplying/Dividing by a Constant
 x x
2
1

2
Constant
Multiple Rule
1
dx   x  1    2 x  dx
 2
2
1
2
    x  1  2 x  dx
 2


2
2


2
1
     g( x )   g '( x )  dx
 2
1  g ( x )

C
2
3
3
1 2

x 1 C
6
3
Integrate


Check
Aim: Integration by Substitution
Course: Calculus
Change of Variable
-u
Recognizing
the Pattern
u 3
1 g(x)
2
y
x 1 4
3


g(x)
u 2
dy
2
 x  1 2x
dx


u
inside function g(x)
g(x)
u
g’(x)
u’
3
2
1 2
2
x 1 C
Evaluate  x  1  2 x  dx 
3




u2 2
(g(x))
Recognizing the f '( g( x )) g '( x ) Pattern
u = g(x) = x2 + 1
u’ = g’(x) = 2x
f ( g ( x ))  ( g( x ))2  f ( u)  f ( x 2  1)  u 2
Aim: Integration by Substitution
Course: Calculus
Change of Variables - u
 f ( g( x )) g '( x ) dx   f (u)du  F (u)  c
Evaluate  2 x  1 dx
Calculate the
differential
du  2dx
u  2x  1
u  2x  1
du
 dx
2
Constant
Multiple Rule
Substitute in
terms of u

1 12
 du 
u     u du
2
 2 
32

1 u 
1 32
Antiderivative
 
C  u C

in terms of u
2 3 2 
3
1
32
Antiderivative

2
x

1

C


in terms of x
3
Aim: Integration by Substitution
Course: Calculus
Guidelines for Making Change of Variables
1. Choose a substitution u = g(x). Usually it
is best to choose the inner part of a
composite function, such as a quantity
raised to a power.
2. Compute du = g’(x) dx
3. Rewrite the integral in terms of the
variable u.
4. Evaluate the resulting integral in terms
of u.
5. Replace u by g(x) to obtain an
antiderivative in terms of x.
6. Check your answer by differentiating.
Aim: Integration by Substitution
Course: Calculus
Model Problem
Evaluate  x 2 x  1 dx
u1
x
2
Substitute in
terms of u
u  2x  1
u  2x  1
du
 dx
2
 u  1  1 2  du 
 
u  

 2 
 2 




1
1
12
   u  1 u
du   u3 2  u1 2 du
4
4
Antiderivative
in terms of u
Antiderivative
in terms of x
52
32

1 u
u 
 

C

4 5 2 3 2 
1
1
52
32
  2 x  1   2 x  1  C
10
6
Aim: Integration by Substitution
Course: Calculus
Model Problem
Evaluate  sin 2 3 x cos3 x dx
u = sin 3x
du
  cos 3 x  3  du   cos 3 x  3  dx
dx
du
  cos 3 x  dx
sin
xdx


cos
x

C
3

cos
ax
2
2 du
sin
axdx



C
sin 3 xacos 3 x dx  u

3


1 2
1  u3 
  u du     C
3 3 
3
u3
1 3

 C  sin 3 x  C
9
9
Check
Aim: Integration by Substitution
Course: Calculus
General Power Rule for Integration
If g is a differentialbe function of x , then
g ( x )

 dx 
  g( x ) g ' x  
n
n 1
C
n1
Equivalently, if u  g( x ), then
n1
u
n
u
 du  n  1  C
n  1
u4
u = 3x – 1
 3  3 x  1
4
n  1
du
dx    3 x  1  3  dx
4
u5/5
3 x  1


5
Aim: Integration by Substitution
5
C
Course: Calculus
Model Problems
u=
x2
+x
  2 x  1  x
2
u1

 x dx    x  x   2 x  1 dx
u2/2
2
 3x
x  2 dx
x

2
C
2
u1/2
u = x3 – 2
3
1
2
x


2
du

du
  x 2
3

12
3 x 2dx
u3 2 /  3 2 
x


Aim: Integration by Substitution
3
2
32

32
C
Course: Calculus
Model Problems

u=1–
4 x
1  2 x 
2
u-2
2x2
du
dx    1  2 x 
u-1/(-1)
2
2
1  2x 


2
C
u2
u = cos x
 cos
 4 x  dx
1
1
2
2
du
x sin x dx     cos x    sin x  dx
2
3
u /3
cos x 


3
Aim: Integration by Substitution
3
C
Course: Calculus
Change of Variables for Definite Integrals
If the function u  f ( x ) has a continuous
derivative on the closed interval [a , b]
and f is continuous on the range of g , then
 f  g  x   g ' x  dx  
b
g(b)
a
g(a )
u = x2 + 1
f  u  du
du = 2x dx
3
1
3
1 1 2
2
0 x  x  1 dx  2 0 x  1  2 x  dx
determine new upper and lower limits of integration

lower limit
When x = 0, u = 02 + 1 = 1
integration
limits for x
and u

upper limit
x = 1, u = 12 + 1 = 2
2
1
1  15
1 u 
1 2 3
  u du      4   
1
2
4 8
2  4 1
2
Aim: Integration by Substitution
Course: Calculus
4
Model Problem
5
x
Evaluate A = 
dx
1
2x  1
u 1
dx  u du
x
u  2x  1
u  2x  1
2
determine new upper and lower limits of integration
lower limit
upper limit
When x = 1, u = 1
x = 5, u = 3
2
2

3
1
2

1 u 1
1 3 2
u du   u  1 du


u 2 
2 1
3
1
1

 16
1 u
   u   9  3   1  
2
3
2 3
 3
1
3
Aim: Integration by Substitution
Course: Calculus
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
Aim: Integration by Substitution
4
u2  1
A =
du
1
2
Area of region
is 16/3
3
Course: Calculus
Even and Odd Functions
Let f be integrable on the closed
interval [  a , a ].
1. If f is an even function, the

a
a
f  x  dx  2  f  x  dx .
a
0
2. If f is an odd function, the

a
a
f  x  dx  
0
a
f  x  dx   f  x  dx .
a
0
To prove even, use f(x) = f(-x) and then
substitute u = -x
Aim: Integration by Substitution
Course: Calculus
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 )
Aim: Integration by Substitution
Course: Calculus
Multiplying/Dividing by a Constant
Aim: Integration by Substitution
Course: Calculus
Multiplying/Dividing by a Constant
Aim: Integration by Substitution
Course: Calculus