LESSON 7 INTEGRATION BY PARTS
Let u and v be differentiable functions of x and consider the differential d ( u v ) .
Since the function u v is a function of x, then by the definition of differential, we
d
d
have that d ( u v )  ( u v ) dx . By the Product Rule, we have that ( u v ) =
dx
dx
du 
dv
du
d
 dv
 v
u
 v . Then d ( u v ) 
( u v ) dx =  u
 dx =
dx 
dx
dx
dx
 dx
dv
du
u
dx  v
dx = u dv  v du . Thus, d ( u v )  u dv  v du . Then we have that
dx
dx
 d ( u v )   u dv   v du . Since  d ( u v )  u v , then we obtain that u v =
 u dv   v du
The equation
 u dv  u v   v du .

 u dv  u v   v du is the formula for evaluating an integral by the
technique of integration called Integration by Parts. To use this formula on an
integral, you must identify the function u and the differential dv.
Example Evaluate
Let
u  x
and
 x cos x dx
dv  cos x dx
v 
Then du  dx
Thus,
 x cos x dx using Integration by Parts.
=
 u dv
 cos x dx  sin x  C
= uv 
 v du
= x ( sin x  C ) 
 (sin x  C ) dx
x sin x  C x  (  cos x  C x )  c = x sin x  C x  cos x  C x  c =
x sin x  cos x  c
Answer: x sin x  cos x  c
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
=
COMMENT: It appears that the constant of integration C that was involved in
integrating the differential dv in order to obtain the function v was not needed for
the example above. Let’s see if this is true in general.
Example Evaluate
Let
 f ( x ) g ( x ) dx using integration by parts.
u  f ( x)
and
v 
Then du  f ( x ) dx
Thus,
 f ( x ) g ( x ) dx
dv  g ( x ) dx
 u dv
=
 g( x ) dx  g ( x )  C
= uv 
 v du
=
f ( x ) [ g ( x )  C ]   [ g ( x )  C ] f ( x ) dx =
f ( x ) g( x )  C f ( x ) 
 [ g ( x ) f ( x )
f ( x ) g( x )  C f ( x ) 
  g( x ) f ( x ) dx  C  f ( x ) dx  =
f ( x ) g( x )  C f ( x ) 
 g ( x ) f ( x ) dx
 C
f ( x ) g( x )  C f ( x ) 
 g ( x ) f ( x ) dx
 C f ( x)  c =
f ( x ) g( x ) 
 g ( x ) f ( x ) dx 
Answer: f ( x ) g ( x ) 
 C f ( x ) ] dx =

f ( x ) dx =
c
 g ( x ) f ( x ) dx 
c
COMMENT: Thus, when we integrate the differential dv in order to obtain the
function v, we do not need to include a constant of integration. This will simplify
our work.
When applying the technique of Integration by Parts, it is helpful to know the
following integral formulas, where k is a constant:
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1
sin k x  c
k
1.
 cos k x dx 
2.
 sin k x dx  
3.
2
 sec k x dx 
4.
2
 csc k x dx  
5.
 sec k x tan k x dx 
6.
 csc k x cot k x dx   k csc k x  c
7.
 tan k x dx 
1
ln sec k x  c
k
8.
 cot k x dx 
1
ln sin k x  c
k
9.
kx
 e dx 
1
cos k x  c
k
1
tan k x  c
k
1
cot k x  c
k
1
sec k x  c
k
1
1 kx
e  c
k
Examples Evaluate the following integrals.
1.
 xe
Let
6x
dx
u  x
Then du  dx
and
dv  e 6 x dx
1
v  e 6x
6
Copyrighted by James D. Anderson, The University of Toledo
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
x e 6 x dx =
 u dv = u v 
 v du =
1 6x 1
x e   e 6 x dx =
6
6
1 6x 1 1 6x
1
1
xe   e  c = xe 6x  e 6x  c
6
6 6
6
36
Answer:
2.
  csc
2
1 6x
1
1
x e  e 6 x  c OR
( 6 x  1) e 6 x  c
6
36
36
 d
u 
Let
Then du  d
  csc
2
 d =
dv  csc 2  d
v   cot 
and
 u dv
= uv 
 v du
=   cot  
 cot  d
=
  cot   ln sin   c
Answer:   cot   ln sin   c
3.

5  / 24
  /6
t sin 4 t dt
There are two ways to apply the Integration by Parts technique to evaluate a
definite integral. The first way is to apply the technique to the indefinite
integral in order to obtain all the antiderivatives of the integrand. Then
evaluate the definite integral using the Fundamental Theorem of Calculus
with one on these antiderivatives. This way is preferable if you have to apply
the Integration by Parts technique more than one time. The second way is to
work with the definite integral and keep the limits of integration in the work.
Using the first method, consider the indefinite integral
Copyrighted by James D. Anderson, The University of Toledo
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 t sin 4 t dt .
ut
Let
and
Then du  dt
 t sin 4 t dt =

dv  sin 4 t dt
1
v   cos 4 t
4
 u dv = u v 
 v du = 
1
1
t cos 4 t   cos 4 t dt =
4
4
1
1 1
1
1
t cos 4 t   sin 4 t  c =  t cos 4 t 
sin 4 t  c =
4
4 4
4
16
1
( sin 4 t  4 t cos 4 t )  c
16
Now, that we have all the antiderivatives of the integrand f ( t )  t sin 4 t , we
can apply the Fundamental Theorem of Calculus to evaluate the definite
integral

5  / 24
  /6
t sin 4 t dt . Note that since the function f ( t )  t sin 4 t is
continuous on its domain of definition, which is the set of all real numbers,
  5 
then it is continuous on the closed interval   ,
. Thus, the
 6 24 
Fundamental Theorem of Calculus can be applied. Thus,

5  / 24
  /6
1
16
 5
5
5 

cos
  sin
 sin
6
6
6


NOTE:
1
16
5  / 24
1

( sin 4 t  4 t cos 4 t ) 
t sin 4 t dt =
=
16
   /6
 2  2
 2   
cos  


  =
3
 3 
 3  
3
5

5

1
  cos  
 sin  ; cos
6
6
2
6
6
2
3

1

 2 
 2 
sin  
   cos  
   sin  
; cos  
3
2
3
2
 3 
 3 
sin
 1 5 
3  
3
2  1   

  

 
    =

  2
2
6
2
3
 2   


 
Copyrighted by James D. Anderson, The University of Toledo
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1
16
 1 5 3

3
3
 
1  1 5 3
 
 
  =


  =
 

12
2
3  
16  2
12
2
3 

 2
5 3
6 3
1  6
4  
6  4   6 3  5



=
16  12
12
12
12 
192
Using the second method, consider the definite integral
ut
Let
Then du  dt

5  / 24
  /6
1

4
and
5  / 24
  /6
t sin 4 t dt .
dv  sin 4 t dt
1
v   cos 4 t
4
5  / 24
1
1


t sin 4 t dt =  t cos 4 t 
4
4
   /6

5  / 24
  /6
cos 4 t dt =
5  / 24
5 
 5
 2   1

cos

cos


sin
4
t


=
 24


6
6
 3   16

  /6
 5

 24

3    1 

       1  sin 5  sin   2   =


2  6  2   16 
6
 3 


1
4

 5
3 
1  5 3
  1  1
 = 1




4 
48
12  16  2
2 
4 
5

3
3  4
1 3
5

=
192
32
Answer:
6  4   6 3  5
192
3  4  1  1  3 


 =


48
2 
 16 
3  4  6  6 3
192
3
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
4.
 w sec
w
w
tan dw
5
5
Let
u  w
and
Then du  dw
 w sec
w
w
tan dw =
5
5
5 w sec
w
w
w
w
w
w
 5  5 ln sec  tan
 c = 5 w sec  25 ln sec  tan
 c
5
5
5
5
5
5
Answer: 5 w sec
5.
w
w
tan dw
5
5
w
v  5 sec
5
dv  sec

6
2
 u dv = u v 
 v du = 5 w sec
w
w
 5  sec dw =
5
5
w
w
w
 25 ln sec  tan
 c
5
5
5
x 2 e  x / 3 dx
First, we will apply the Integration by Parts technique to evaluate the
2  x/3
dx . Then we will choose one of these
indefinite integral  x e
antiderivatives to apply the Fundamental Theorem of Calculus to find the
value of the given definite integral.
u  x2
Let
and
Then du  2 x dx
x
2
e  x / 3 dx =
 u dv
dv  e  x / 3 dx
v   3e  x / 3
= uv 
 v du
2  x/3
 6  x e  x / 3 dx
=  3x e
Now, apply the Integration by Parts technique again in order to evaluate the
 x/3
dx .
indefinite integral  x e
u  x
Let
Then du  dx
and
dv  e  x / 3 dx
v   3e  x / 3
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 xe
 x/3
dx =
 u dv
= uv 
 v du
 x/3
 3  e  x / 3 dx =
=  3xe
 3 x e  x / 3  3(  3 ) e  x / 3  c =  3 x e  x / 3  9 e  x / 3  c
Thus, we have that
x
2
e  x / 3 dx =  3 x 2 e  x / 3  6  x e  x / 3 dx =
 3 x2 e  x / 3  6(  3 x e  x / 3  9 e  x / 3 )  c =
 3 x 2 e  x / 3  18 x e  x / 3  54 e  x / 3  c =
 3( x 2  6 x  18 ) e  x / 3  c
2  x/3
Now, that we have all the antiderivatives of the integrand f ( x )  x e
, we
can apply the Fundamental Theorem of Calculus to evaluate the definite
integral

6
2
x 2 e  x / 3 dx . Note that since the function f ( x )  x 2 e  x / 3 is
continuous on its domain of definition, which is the set of all real numbers,
then it is continuous on the closed interval [  2 , 6 ] . Thus, the Fundamental
Theorem of Calculus can be applied. Thus,

6
2
x 2 e  x / 3 dx =  3 ( x 2  6 x  18 ) e  x / 3

6
2
=
 3 [ ( 36  36  18 ) e  2  ( 4  12  18 ) e 2 / 3 ] =  3 ( 90 e  2  10 e 2 / 3 ) =
3 (10 e
2/3
 90 e
2
) = 30 e
2
(e
8/3
30 ( e 8 / 3  9 )
 9) =
e2
30 ( e 8 / 3  9 )
Answer:
e2
NOTE: There is a faster way to apply the Integration by Parts technique to
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
evaluate the indefinite integral
x
2
e  x / 3 dx . We will discuss it later in this
lesson.
6.
e
2t
Let
cos 7 t dt
u  e 2t
and
2t
Then du  2 e dt
2t
 e cos 7 t dt =
dv  cos 7 t dt
1
v  sin 7 t
7
 u dv = u v 
 v du =
1 2t
2
e sin 7 t   e 2 t sin 7 t dt
7
7
Now, apply the Integration by Parts technique again in order to evaluate the
2t
indefinite integral  e sin 7 t dt .
Let
u  e 2t
and
2t
Then du  2 e dt
2t
 e sin 7 t dt =
dv  sin 7 t dt
1
v   cos 7 t
7
 u dv = u v 
 v du = 
1 2t
2
e cos 7 t   e 2 t cos 7 t dt
7
7
Thus, we have that
2t
 e cos 7 t dt =
1 2t
2
e sin 7 t 
7
7
1 2t
2
e sin 7 t   e 2 t sin 7 t dt =
7
7
2
 1 2t


e
cos
7
t

e 2 t cos 7 t dt  =

 7
7

1 2t
2 2t
4
e sin 7 t 
e cos 7 t 
e 2 t cos 7 t dt

7
49
49
Thus, we have that
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2t
 e cos 7 t dt =
1 2t
2 2t
4
e sin 7 t 
e cos 7 t 
e 2 t cos 7 t dt

7
49
49
Notice that the indefinite integral
e
2t
cos 7 t dt , which we are evaluating,
appears on both sides of the equation above. Thus, we could solve for
2t
 e cos 7 t dt .
Adding
4
e 2 t cos 7 t dt to both sides of the equation, we obtain that

49
53
1 2t
2 2t
2t
e
cos
7
t
dt
e
sin
7
t

e cos 7 t 
=
49 
7
49
where C is a constant, then we have that
 0 dt .
Since
 0 dt  C ,
53
e 2 t cos 7 t dt =

49
1 2t
2 2t
e sin 7 t 
e cos 7 t  C . Now, multiplying both sides of this equation
7
49
49
, we obtain that
53
by
2t
 e cos 7 t dt =
7 2t
2 2t
49
e sin 7 t 
e cos 7 t 
C =
53
53
53
7 2t
2 2t
1 2t
e sin 7 t 
e cos 7 t  c =
e ( 7 sin 7 t  2 cos 7 t )  c , where
53
53
53
c 
49
C.
53
Answer:
7.

4
2
1 2t
e ( 7 sin 7 t  2 cos 7 t )  c
53
ln x dx
Copyrighted by James D. Anderson, The University of Toledo
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First, we will apply the Integration by Parts technique to evaluate the
indefinite integral  ln x dx . Then we will choose one of these
antiderivatives to apply the Fundamental Theorem of Calculus to find the
value of the given definite integral.
u  ln x and
1
Then du  dx
x
Let
 ln x dx
=
 u dv
dv  dx
v  x
= uv 
 v du
= x ln x 
 dx
= x ln x  x  c
Now, that we have all the antiderivatives of the integrand f ( x )  ln x , we
can apply the Fundamental Theorem of Calculus to evaluate the definite
integral

4
2
ln x dx . Note that since the function f ( x )  ln x is continuous
on its domain of definition, which is the set of all positive real numbers, then
it is continuous on the closed interval [ 2 , 4 ] . Thus, the Fundamental
Theorem of Calculus can be applied. Thus,

4
2
ln x dx = x ln x  x  42 = 4 ln 4  4  ( 2 ln 2  2 ) = ln 4 4  4  ln 4  2 =
44
ln
 2 = ln 4 3  2 = ln 64  2
4
Or, you could have applied the Integration by Parts technique to evaluate the
definite integral

4
2
ln x dx in the following manner.
u  ln x and
1
Then du  dx
x
Let

4
2
ln x dx = x ln x  2 
4
dv  dx
v  x

4
2
dx = 4 ln 4  2 ln 2  x  42 =
44
 2 = ln 4 3  2 = ln 64  2
ln 4  ln 4  ( 4  2 ) = ln
4
4
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Answer: ln 64  2
8.
 Arc cos 3t dt
dv  dt
u  Arc cos 3 t
and
3
du


dt
Then
1  9t 2
Let

 u dv = u v 
Arc cos 3t dt =
The indefinite integral

vt
 v du = t Arc cos 3t  3 
t
1  9t
2
 18 t
1
18

1
1  9t 2  c
9

1  9t 2
dt = 
1  9t
2
dt
dt can be evaluated by simple
2
substitution. Let w  1  9 t . Then dw   18 t dt . Then

t

t
1  9t 2
dt =
1 du
1
1
 1/ 2

u
du

 2 u 1/ 2  c =
=
=


18
18
18
u
Thus, we have that

Arc cos 3t dt = t Arc cos 3 t  3 
t
1  9t
2
dt =
1
 1

t Arc cos 3 t  3  
1  9 t 2   c = t Arc cos 3 t 
1  9t 2  c =
3
 9

1
( 3 t Arc cos 3 t 
3
1  9t 2 )  c
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
Answer:
9.

3
1
( 3 t Arc cos 3 t 
3
x ln x dx
u  ln x and
1
Then du  dx
x
Let

3
1  9t 2 )  c
x ln x dx =
dv  x 1 / 3dx
v 
3 4/3
x
4
 u dv = u v 
 v du =
3 4/3
3
x ln x 
4
4
x
1/ 3
dx =
3 4/3
3 3
3 4/3
9 4/3
x ln x   x 4 / 3  c =
x ln x 
x  c =
4
4 4
4
16
3 4/3
x ( 4 ln x  3 )  c
16
Answer:
10.
3 4/3
x ( 4 ln x  3 )  c
16

ln y 5
dy
8y

ln y 5
dy =
8y

5 ln y
5
dy =
8y
8
The indefinite integral

Let u  ln y . Then du 

ln y
dy
y
ln y
dy can be evaluated by simple substitution.
y
1
dy .
y
Copyrighted by James D. Anderson, The University of Toledo
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
ln y 5
5
dy =
8y
8
Answer:
11.
x
n

ln y
5
dy =
y
8
5
( ln y ) 2  c
16
ln x dx , where n is a rational number such that n   1
Let
u  ln x
dv  x n dx
and
xn  1
v 
n 1
1
Then du  dx
x

x n ln x dx =
1
xn
n 1
 1
1
xn
2
( n  1)
Answer:
12.
 x tan
Let

5
5 u2
u du = 
( ln y ) 2  c
 c =
16
8 2
1
 u dv = u v 
 v du =
1
xn
n 1
1
1
xn  1
xn
ln x 

 c =
n 1
n 1 n 1
 1
 1
 1
ln x 
[ ( n  1 ) ln x  1 ]  c
1
xn
2
( n  1)
 1
[ ( n  1 ) ln x  1 ]  c
x dx
u  tan  1 x
and
1
dx
Then du 
1  x2
ln x 
dv  x dx
x2
v 
2
Copyrighted by James D. Anderson, The University of Toledo
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1
x n dx =

n 1
1
xn
2
( n  1)
 1
 c =
 x tan
1
1 2 1
1
x2
x tan x  
dx
x dx =  u dv = u v   v du =
2
2 1  x2
x2
dx can be evaluated by rewriting the
The indefinite integral 
1  x2
integrand in the following manner.
x2
 1  x 2 dx =


x2
dx =
x2  1

x2  1  1
dx =
x2  1
 x2  1
1


  x 2  1 x 2 

 dx =
1 

1 
1  2
 dx = x  tan  1 x  c
x  1

Thus, we have that
 x tan
1
1 2 1
1
x2
x tan x  
dx =
x dx =
2
2 1  x2
1 2 1
1
1
x tan x  ( x  tan  1 x )  c = ( x 2 tan  1 x  x  tan  1 x )  c
2
2
2
Answer:
13.
 ln
Let
1 2 1
( x tan x  x  tan  1 x )  c
2
7 y  5 dy
u  ln 7 y  5
Then du 
7
dy
7y  5
 ln 7 y  5 dy =
and
dv  dy
v  y
 u dv = u v 
 v du = y ln 7 y  5 
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860

7y
dy
7y  5
7y
 7 y  5 dy can be evaluated by rewriting the
integrand in the following manner.
The indefinite integral
7y  5  5
dy =
7y  5
 7y  5
5 


  7 y  5 7 y  5  dy =

7y
dy =
7y  5



5
5
1 
 dy = y  ln 7 y  5  c
7y  5 
7


Thus, we have that
5


y
ln
7
y

5

y

ln
7
y

5
ln
7
y

5
dy

  c
=

7


y ln 7 y  5  y 
5
ln 7 y  5  c =
7
1
( 7 y ln 7 y  5  7 y  5 ln 7 y  5 )  c =
7
1
[ ( 7 y  5 ) ln 7 y  5  7 y ]  c
7
Answer:
14.
t
4
1
[ ( 7 y  5 ) ln 7 y  5  7 y ]  c
7
t
cot  1 dt
4
t
4
4
dt
Then du  
16  t 2
Let
u  cot  1
and
dv  t 4dt
t5
v 
5
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1
1
1
4

1 t 




D
cot
2
2


NOTE: t
=
=
=
t 4
t
16  t 2
4

1
4 
16
4
t
4
cot
1
t
dt =
4
1 5
4
1
t
cot
t

u
dv
u
v

v
du
=
=


5
5
The indefinite integral


t5
dt
t 2  16
t5
dt can be evaluated by rewriting the
t 2  16
integrand.
5
2
First, carry out the long division of t by t  16 :
t 3  16 t
t 2  16 t 5
t 5  16 t 3
 16 t 3
 16 t 3  256 t
256 t
t5
256 t
 t 3  16 t  2
Thus, 2
. Then
t  16
t  16

t5
dt =
t 2  16
 3
2t 
t4
2
2
  t  16 t  128  t 2  16  dt = 4  8 t  128 ln ( t  16 )  c
Thus, we have that
t
4
cot
1
1
4
t
dt = t 5 cot  1 t 
5
5
4

t5
dt =
t 2  16

1 5 1
4 t4
t cot t    8 t 2  128 ln ( t 2  16 )   c =
5
54

Copyrighted by James D. Anderson, The University of Toledo
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t4

1  5 1
2
2
t
cot
t

4

8
t

128
ln
(
t

16
)

 4
  c =
5


1 5 1
[ t cot t  t 4  32 t 2  512 ln ( t 2  16 ) ]  c
5
Answer:
1 5 1
[ t cot t  t 4  32 t 2  512 ln ( t 2  16 ) ]  c
5
Let return to Example 5, where we applied the Integration by Parts technique two
2  x/3
dx to obtain that  x 2 e  x / 3 dx =
times on the indefinite integral  x e
 3( x 2  6 x  18 ) e  x / 3  c . Since the expression x 2 can be differentiated down to
0, we could set up the following table to help us carry out the repeated Integration
by Parts:
D
I
x2
e  x/3
2x
 3e  x / 3
+
 x/3
add the product of x 2 and  3 e
,
2
9 e  x/3

 x/3
subtract the product of 2 x and 9 e
,
0
 27 e  x / 3
+
 x/3
add the product of 2 and  27 e
,
To obtain the value of the indefinite integral,
and add the constant of integration.
Thus, we have that
x
2
e  x / 3 dx =  3 x 2 e  x / 3  18 x e  x / 3  54 e  x / 3  c =
 3( x 2  6 x  18 ) e  x / 3  c .
NOTE: You are NOT allowed to use this table method for one application of the
Integration by Parts technique.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
15.
x
3
cos 2 x dx
Since the expression x 3 can be differentiated down to 0, we could set up the
following table to help us carry out the repeated Integration by Parts:
D
I
x3
cos 2 x
3x 2
To obtain the value of the indefinite integral,
sin 2 x
+
add the product of x 3 and
6x
 14 cos 2 x

2
subtract the product of 3x and  14 cos 2 x ,
6
 81 sin 2 x
+
add the product of 6 x and  81 sin 2 x ,

subtract the product of 6 and
1
2
1
16
0
cos 2 x
1
2
sin 2 x ,
1
16
cos 2 x ,
and add the constant of integration.
Thus, we have that
x
3
cos 2 x dx =
1 3
3
3
3
x sin 2 x  x 2 cos 2 x  x sin 2 x  cos 2 x  c =
2
4
4
8
1
( 4 x 3 sin 2 x  6 x 2 cos 2 x  6 x sin 2 x  3 cos 2 x )  c
8
Answer:
16.
y
4
1
( 4 x 3 sin 2 x  6 x 2 cos 2 x  6 x sin 2 x  3 cos 2 x )  c
8
e 5 y dy
4
Since the expression y can be differentiated down to 0, we could set up the
following table to help us carry out the repeated Integration by Parts:
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
D
I
y4
e5y
4y3
1
5
12 y 2
To obtain the value of the indefinite integral,
e5y
+
4
add the product of y and
1
25
e5y

3
subtract the product of 4 y and
24 y
1
125
e5y
+
2
add the product of 12 y and
24
1
625
e5y

subtract the product of 24 y and
+
add the product of 24 and
0
1
3125
e5y
1
5
e5y,
1
3125
1
125
1
25
e5y ,
e5y ,
1
625
e5y ,
e5y ,
and add the constant of integration.
Thus, we have that
y
4
e 5 y dy =
1 4 5y
4 3 5y
12 2 5 y
24
24 5 y
y e 
y e 
y e 
ye5y 
e  c =
5
25
125
625
3125
1 5y
e ( 625 y 4  500 y 3  300 y 2  120 y  24 )  c
3125
Answer:
17.

1 5y
e ( 625 y 4  500 y 3  300 y 2  120 y  24 )  c
3125
2
x e x dx
2
Since D x x  2 x and we have the x in the integral, then this indefinite
integral can be evaluated by simple substitution.
2
Let u  x . Then du  2 x dx .
Copyrighted by James D. Anderson, The University of Toledo
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
Answer:
18.

1
2
2
x e x dx =

2
2 x e x dx =
1
2

e u du =
1 u
1 2
e  c = ex  c
2
2
1 x2
e  c
2
3
x 5e x dx
We will use the Integration by Parts technique on this integral. Note that you
3
2 x3
can not integrate e x . However, the expression x e can be integrated by
simple substitution. Thus,
u  x3
Let
and
2
Then du  3 x dx
To find
Then



dv  x 2 e x dx
1 3
v  ex
3
3
3
x 2 e x dx : Let w  x 3 . Then dw  3 x 2 dx .
3
x 2 e x dx =
3
x 5e x dx =
1
1 w
1 w
1 x3
2 x3
3
x
e
dx
e
dw
e

c
= 
=
= e  c
3
3
3
3
 u dv = u v 
 v du =
1 3 x3
x e 
3

1 3 x3 1 x3
1 3
x e  e  c = e x ( x 3  1)  c
3
3
3
Answer:
19.
t
3
1 x3 3
e ( x  1)  c
3
cos 3t 2 dt
Copyrighted by James D. Anderson, The University of Toledo
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3
x 2 e x dx =
We will use the Integration by Parts technique on this integral. Note that you
2
2
can not integrate cos 3t . However, the expression t cos 3t can be
integrated by simple substitution. Thus,
u  t2
Let
and
Then du  2 t dt
To find
Then
 t cos 3t
2
dv  t cos 3t 2 dt
1
v  sin 3 t 2
6
dt : Let w  3t 2 . Then dw  6 t dt .
2
 t cos 3t dt =
1
1
1
2
6
t
cos
3
t
dt
cos
u
du
sin u  c =
=
=
6
6
6
1
sin 3 t 2  c
6
3
2
 t cos 3t dt =
 u dv = u v 
 v du =
1 2
1
t sin 3 t 2   t sin 3 t 2 dt =
6
3
1 2
1 1
1
1

t sin 3 t 2    cos 3 t 2   c = t 2 sin 3 t 2 
cos 3 t 2  c =
6
3 6
6
18

1
( 3 t 2 sin 3 t 2  cos 3 t 2 )  c
18
Answer:
20.
 csc
3
1
( 3 t 2 sin 3 t 2  cos 3 t 2 )  c
18
6 d
u  csc 6
Let
and
Then du   6 csc 6 cot 6 d
 csc
3
6 d =
 u dv
= uv 
 v du
dv  csc 2 6 d
1
v   cot 6
6
=
Copyrighted by James D. Anderson, The University of Toledo
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
1
csc 6 cot 6 
6
 csc 6 cot
2
6 d =
2
2
2
2
NOTE: csc 6  cot 6  1  cot 6  csc 6  1

1
csc 6 cot 6 
6
 csc 6 ( csc

1
csc 6 cot 6 
6
 ( csc

1
csc 6 cot 6 
6
  csc 6 d   csc 6 d  =

1
csc 6 cot 6 
6
 csc

1
csc 6 cot 6 
6
 csc
3
2
6  1) d =
6  csc 6 ) d =
3
3
6 d 
 csc 6 d
3
6 d 
1
ln csc 6  cot 6
6
=
Thus, we have that
3
 csc 6 d = 
1
csc 6 cot 6 
6
Notice that the indefinite integral
3
 csc 6 d 
 csc
3
1
ln csc 6  cot 6
6
6 d , which we are evaluating,
appears on both sides of the equation above. Thus, we could solve for
3
 csc 6 d .
Adding
 csc
3
6 d to both sides of the equation, we obtain that
2  csc 3 6 d = 
Since
 0 d
1
1
csc 6 cot 6  ln csc 6  cot 6 
6
6
 0 d .
 C , where C is a constant, then we have that 2  csc 3 6 d =
Copyrighted by James D. Anderson, The University of Toledo
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1
1
csc 6 cot 6  ln csc 6  cot 6  C . Now, dividing both sides of
6
6
this equation by 2, we obtain that

3
 csc 6 d = 

1
1
C
csc 6 cot 6 
ln csc 6  cot 6 
=
12
12
2
1
1
csc 6 cot 6 
ln csc 6  cot 6  c =
12
12
1
C
( ln csc 6  cot 6  csc 6 cot 6 )  c , where c  .
12
2
1
( ln csc 6  cot 6  csc 6 cot 6 )  c
Answer:
12
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