Warm-up:
Evaluate the integrals.
 x 7
1)   3e  dx
x

2)
(
x
1
3 1 x
2
)dx
Warm-up:
Evaluate the integrals.
 x 7
x
1)   3e  dx  3e  7 ln x  C
x

2)
(
x
1
3 1 x
2
)dx
Warm-up:
Evaluate the integrals.
 x 7
x
1)   3e  dx  3e  7 ln x  C
x

2)
( x 
1
3
2
1
)dx  2 x  sin x  C
2
3 1 x
3
3
Integration by Parts
Section 8.2
Objective: To integrate problems
without a u-substitution
Integration by Parts
• When integrating the product of two functions, we
often use a u-substitution to make the problem easier
to integrate. Sometimes this is not possible. We need
another way to solve such problems.
 f ( x)  g ( x)
Integration by Parts
• As a first step, we will take the derivative of f ( x)  g ( x)
Integration by Parts
• As a first step, we will take the derivative of f ( x)  g ( x)
d
 f ( x)  g ( x)  f ( x)  g / ( x)  g ( x)  f / ( x)
dx
Integration by Parts
• As a first step, we will take the derivative of f ( x)  g ( x)
d
 f ( x)  g ( x)  f ( x)  g / ( x)  g ( x)  f / ( x)
dx

d
 f ( x)  g ( x)   f ( x)  g / ( x)   g ( x)  f / ( x)
dx
Integration by Parts
• As a first step, we will take the derivative of f ( x)  g ( x)
d
 f ( x)  g ( x)  f ( x)  g / ( x)  g ( x)  f / ( x)
dx

d
 f ( x)  g ( x)   f ( x)  g / ( x)   g ( x)  f / ( x)
dx
 f ( x)  g ( x)  
f ( x)  g / ( x)   g ( x)  f / ( x)
Integration by Parts
• As a first step, we will take the derivative of f ( x)  g ( x)
d
 f ( x)  g ( x)  f ( x)  g / ( x)  g ( x)  f / ( x)
dx

d
 f ( x)  g ( x)   f ( x)  g / ( x)   g ( x)  f / ( x)
dx
 f ( x)  g ( x)  
f ( x)  g / ( x)   g ( x)  f / ( x)
 f ( x)  g ( x)   g ( x)  f / ( x)  
f ( x)  g / ( x)
Integration by Parts
• Now lets make some substitutions to make this
easier to apply.
u  f (x)
v  g (x)
du  f / ( x)
dv  g / ( x)
 f ( x) g ( x)   g ( x) f / ( x)  
uv   vdu   udv
f ( x) g / ( x)
Integration by Parts
• This is the way we will look at these problems.
u  f (x)
du  f / ( x)
v  g (x)
dv  g / ( x)
uv   vdu   udv
• The two functions in the original problem we are
integrating are u and dv. The first thing we will do is
to choose one function for u and the other function
will be dv.
Example 1
• Use integration by parts to evaluate
 x cos xdx
Example 1
• Use integration by parts to evaluate
ux
dv  cos xdx
 x cos xdx
Example 1
• Use integration by parts to evaluate
ux
dv  cos xdx
du  dx
v  sin x
 x cos xdx
Example 1
• Use integration by parts to evaluate
ux
dv  cos xdx
du  dx
v  sin x
 x cos xdx
 x cos xdx x sin x   sin xdx
Example 1
• Use integration by parts to evaluate
ux
dv  cos xdx
du  dx
v  sin x
 x cos xdx
 x cos xdx x sin x   sin xdx
 x cos xdx x sin x  cos x  C
Guidelines
• The first step in integration by parts is to choose u
and dv to obtain a new integral that is easier to
evaluate than the original. In general, there are no
hard and fast rules for doing this; it is mainly a
matter of experience that comes from lots of
practice.
Guidelines
• There is a useful strategy that may help when
choosing u and dv. When the integrand is a product
of two functions from different categories in the
following list , you should make u the function whose
category occurs earlier in the list.
• Logarithmic, Inverse Trig, Algebraic, Trig, Exponential
• The acronym LIATE may help you remember the order.
Guidelines
• If the new integral is harder that the original, you
made the wrong choice. Look at what happens when
we make different choices for u and dv in example 1.
Guidelines
• If the new integral is harder that the original, you
made the wrong choice. Look at what happens when
we make different choices for u and dv in example 1.
 x cos xdx
u  cos x
du   sin xdx
x2
v
2
dv  xdx
x2
x2
 x cos xdx  2 cos x   2 sin xdx
Guidelines
• Since the new integral is harder than the original, we
made the wrong choice.
 x cos xdx
u  cos x
du   sin xdx
x2
v
2
dv  xdx
x2
x2
 x cos xdx  2 cos x   2 sin xdx
Example 2
• Use integration by parts to evaluate
x
xe
 dx
Example 2
• Use integration by parts to evaluate
ux
dv  e x dx
x
xe
 dx
Example 2
• Use integration by parts to evaluate
ux
du  dx
dv  e x dx
v  ex
x
xe
 dx
Example 2
• Use integration by parts to evaluate
ux
du  dx
dv  e x dx
v  ex
x
x
x
xe
dx

xe

e

 dx
x
xe
 dx
Example 2
• Use integration by parts to evaluate
ux
du  dx
dv  e x dx
v  ex
x
x
x
xe
dx

xe

e

 dx
x
x
x
xe
dx

xe

e
C

x
xe
 dx
Example 3 (S):
• Use integration by parts to evaluate
 ln xdx
Example 3
• Use integration by parts to evaluate
u  ln x
dv  dx
 ln xdx
Example 3
• Use integration by parts to evaluate
u  ln x
dv  dx
1
du  dx
x
vx
 ln xdx
Example 3
• Use integration by parts to evaluate
u  ln x
dv  dx
1
du  dx
x
vx
 ln xdx x ln x   dx
 ln xdx
Example 3
• Use integration by parts to evaluate
u  ln x
dv  dx
1
du  dx
x
vx
 ln xdx x ln x   dx
 ln xdx x ln x  x  C
 ln xdx
Example 4 (Repeated):
• Use integration by parts to evaluate
2 x
x
 e dx
Example 4 (Repeated):
• Use integration by parts to evaluate
u  x2
dv  e  x dx
2 x
x
 e dx
Example 4 (Repeated):
• Use integration by parts to evaluate
u  x2
dv  e  x dx
du  2xdx
v  e  x
2 x
x
 e dx
Example 4 (Repeated):
• Use integration by parts to evaluate
u  x2
dv  e  x dx
du  2xdx
v  e  x
2 x
2 x
x
x
e
dx


x
e

2
xe

 dx
2 x
x
 e dx
Example 4 (Repeated):
• Use integration by parts to evaluate
u  x2
dv  e  x dx
du  2xdx
v  e  x
2 x
2 x
x
x
e
dx


x
e

2
xe

 dx
2 x
x
 e dx
ux
dv  e  x dx
Example 4 (Repeated):
• Use integration by parts to evaluate
u  x2
dv  e  x dx
du  2xdx
v  e  x
2 x
2 x
x
x
e
dx


x
e

2
xe

 dx
2 x
x
 e dx
ux
du  dx
dv  e  x dx
v  e  x
Example 4 (Repeated):
• Use integration by parts to evaluate
u  x2
dv  e  x dx
du  2xdx
v  e  x
2 x
2 x
x
x
e
dx


x
e

2
xe

 dx
x e
2 x

dx   x e  2  xe   e dx
2 x
x
x

2 x
x
 e dx
ux
du  dx
dv  e  x dx
v  e  x
Example 4 (Repeated):
• Use integration by parts to evaluate
u  x2
dv  e  x dx
du  2xdx
v  e  x
2 x
2 x
x
x
e
dx


x
e

2
xe

 dx
x e
2 x

dx   x e  2  xe   e dx
2 x
x
x

2 x
x
 e dx
ux
du  dx
2 x
2 x
x
x
x
e
dx


x
e

2
xe

2
e

C

dv  e  x dx
v  e  x
Example 5:
1
• Evaluate the following definite integral  (tan 1 x)dx
0
Example 5:
1
• Evaluate the following definite integral  (tan 1 x)dx
0
u  tan 1 x
dv  dx
Example 5:
1
• Evaluate the following definite integral  (tan 1 x)dx
1
u  tan x
1
du 
1 x2
0
dv  dx
vx
Example 5:
1
• Evaluate the following definite integral  (tan 1 x)dx
1
u  tan x
1
1
du 
1 x2
0
dv  dx
xdx
0 (tan x)dx  x tan x   1  x 2
1
1
vx
Example 5:
1
• Evaluate the following definite integral  (tan 1 x)dx
1
u  tan x
1
1
du 
1 x2
0
dv  dx
xdx
1
1
0 (tan x)dx  x tan x   1  x 2
vx
u  1 x2
Example 5:
1
• Evaluate the following definite integral  (tan 1 x)dx
1
u  tan x
1
1
du 
1 x2
0
dv  dx
xdx
1
1
0 (tan x)dx  x tan x   1  x 2
vx
u  1 x2
du  2xdx
Example 5:
1
• Evaluate the following definite integral  (tan 1 x)dx
1
u  tan x
1
1
du 
1 x2
0
dv  dx
xdx
1
1
0 (tan x)dx  x tan x   1  x 2
vx
u  1 x2
du  2xdx
du
 dx
2x
Example 5:
1
• Evaluate the following definite integral  (tan 1 x)dx
1
u  tan x
1
du 
1 x2
0
dv  dx
1
xdx
1
1
0 (tan x)dx  x tan x   1  x 2
1
1 du
0 (tan x)dx  x tan x  2  u
1
1
vx
u  1 x2
du  2xdx
du
 dx
2x
Example 5:
1
• Evaluate the following definite integral  (tan 1 x)dx
1
u  tan x
1
du 
1 x2
0
dv  dx
1
u  1 x2
xdx
1
1
0 (tan x)dx  x tan x   1  x 2
1
du  2xdx
du
 dx
2x
1 du
0 (tan x)dx  x tan x  2  u
1
1
1
1
2
(tan
x
)
dx

x
tan
x

ln(
1

x
)
0
2
1
1
vx

1
0
Example 5:
1
• Evaluate the following definite integral  (tan 1 x)dx
0
1
1
2
(tan
x
)
dx

x
tan
x

ln(
1

x
)
0
2
1
1
10
Example 5:
1
• Evaluate the following definite integral  (tan 1 x)dx
0
1
1
2
(tan
x
)
dx

x
tan
x

ln(
1

x
)
0
2
1
1
1
10
1
1
2
1
2
(tan
x
)
dx

1
tan
1

ln(
1

1
)

0
tan
0

ln(
1

0
)
0
2
2
1
1
Example 5:
1
• Evaluate the following definite integral  (tan 1 x)dx
0
1
1
2
(tan
x
)
dx

x
tan
x

ln(
1

x
)
0
2
1
1
1
1
1
2
1
2
(tan
x
)
dx

1
tan
1

ln(
1

1
)

0
tan
0

ln(
1

0
)
0
2
2
1
1
1

1
0 (tan x)dx  4  2 ln 2  0  0
1
Example 5:
1
• Evaluate the following definite integral  (tan 1 x)dx
0
1
1
2
(tan
x
)
dx

x
tan
x

ln(
1

x
)
0
2
1
1
1
1
1
2
1
2
(tan
x
)
dx

1
tan
1

ln(
1

1
)

0
tan
0

ln(
1

0
)
0
2
2
1
1
1

0 (tan x)dx  4  2 ln 2  0  0  4  ln 2
1
1

Homework:
Page 520
# 3-9 odd, 15, 25, 29,
31, 37