Integration by Parts
I. Basic Technique
By the Product Rule for Derivatives,
d
 f ( x) g ( x)  f ( x) g ( x)  g ( x) f ( x) . Thus,
dx
  f (x)g (x)  g(x) f (x) dx  f (x)g(x)   f (x)g (x) dx   g(x) f (x) dx 
f ( x) g ( x)  f ( x) g ( x) dx  f ( x) g ( x)  g ( x) f ( x) dx . This formula for integration


by parts often makes it possible to reduce a complicated integral involving a product to
a simpler integral. By letting u  f ( x)  du  f ( x)dx
dv  g ( x)dx  v  g ( x)
we get the more common formula for integration by parts:
 udv  uv   vdu
Example 1: Find
 x ln xdx .
1
dx and v 
x

1 2
x . Thus,
2
1 
x ln xdx  (ln x)( x dx)  udv  uv  vdu  (ln x) x 2  
2 
1
1
11 
 1 2  1  1 2
x dx  x 2 ln x   x 2   C 
 x  dx   x ln x 
2
2
22 
 2  x  2
1 2
1
x ln x  x 2  C .
2
4
Let u  ln x and dv  xdx  du 





x dx 

It is possible that when you set up an integral using integration by parts, the resulting
integral will be more complicated than the original integral. In this case, change your
substitutions for u and dv.
1
Example 2: Find
 x sin x dx .

1 2
x . Thus,
2
1 
udv  uv  vdu  (sin x) x 2  
2 
Let u  sin x and dv  x dx  du  cos x dx and v 
x dx 
 x sin x dx   (sin x)( x dx)  

1
1 
  2 x (cos x dx)  2  x cos x dx . Notice that this resulting integral is
2
2
more complicated that the original one. Therefore, let u  x and dv 
 sin x dx   cos x . Thus,  x sin x dx =
 udv  uv   vdu  (x)( cos x)   ( cos x) dx  x cos x   cos xdx 
sin x dx  du  dx and v 
 x cos x  sin x  C .
Example 3: Find
 ln x dx .

1
dx and v  1dx  x . Thus,
x
1 
udv  uv  vdu  (ln x)( x)   dx ( x) 
x 
Let u  ln x and dv  1dx  du 
 ln x dx   (ln x)(1dx)  
x ln x  1dx  x ln x  x  C .

Example 4: Find


 arctan x dx .
As in the previous example, let u  arctan x and dv  dx  du 

 arctan x dx   udv  uv   vdu  x arctan x 
x
1
2x

dx   x arctan x 
dx  x arctan x 
dx 
1 x
2 1 x

and v  1dx  x . Thus,

 1
x
2
1 x
1
dx
1 x2
2
1
1
x arctan x  ln x 2  1  C = x arctan x  ln( x 2  1)  C .
2
2
2
2
Example 5: Find

ln x
dx .
x2

1
1
dx  x  2 dx  du  dx and v  x 2 dx  1x 1 
2
x
x
1
ln x
 1

 1
dx  (ln x) 2 dx   udv  uv  vdu  (ln x)  
. Thus,
2
x
x
x

 x 
1
 ln x
 ln x 1
 1   1   ln x

dx 
 x  2 dx 
 C 
 dx   
2
x
x
x
x
x
 x  x 
 ln x  1
C.
x
Let u  ln x and dv 







Sometimes, it is necessary to use integration by parts more than once.
Example 6: Find

x 2 cos x dx .
Let u  x 2 and dv  cos x dx  du  2 xdx and v 
 cos x dx  sin x . Thus,
 x cos x dx   udv  uv   vdu  (x )(sin x)   (2xdx)(sin x)  x sin x 
2 x sin xdx . Notice that the resulting integral x sin x dx is less complicated


2
2
2
than the original one, but integration by parts is needed to evaluate it.
Let u  x and dv  sin x dx  du  dx and v 
 sin x dx   cos x . Thus,
 x sin x dx   udv  uv   vdu  x( cos x)   ( cos x) dx  x cos x 
 cos x dx  x cos x  sin x . Finally, we get  x cos x dx  x sin x 
2
2
2 x cos x  sin x  C  x 2 sin x  2 x cos x  2 sin x  C . [Note: We will be
able to integrate
 x cos x dx more easily with tabular integration; this
2
technique will be described later.]
The next illustration of repeated integration by parts deserves special attention.
Example 7: Find
 e sin x dx .
x
3
 sin xdx   cos x . Thus,
 e sin x dx   udv  uv   vdu  (e )( cos x)   ( cos x)(e dx) 
 e cos x  e cos xdx . Notice that integration by parts is now needed to

evaluate e cos xdx . Let u  e and dv  cos xdx  du  e dx and

v  cos xdx  sin x . Thus, e cos xdx  udv  uv  vdu  e sin x 




 (sin x)(e dx)  e sin x   e sin xdx . Returning to the original problem,
 e sin x dx =  e cos x   e cos xdx  e cos x  e sin x   e sin x dx .
1
Thus, 2 e sin x dx  e cos x  e sin x  e sin x dx =  e cos x 


2
Let u  e x and dv  sin x dx  du  e x dx and v 
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
1 x
e sin x  C and this does check.
2
The next example illustrates an interesting type of integral that surprisingly requires
integration by parts.
Example 8:
 sin  x  dx .
Let u  x  u 2  x  2udu  dx 
 sin  x  dx   (sin u)(2udu) 

 x sin x dx =  x cos x  sin x  C or  u sin u du =  u cos u  sin u  C .
Thus, sin  x  dx = 2 u sin u du =  2u cos u  2 sin u  C  2 x cos x  


2 u sin u du . In example 2, we got the following using integration by parts:
2 sin
 x   C and this does check.
In general, to evaluate

f
 x  dx = n u
n
n 1
 f  x  dx , let u 
n
f (u ) du .
4
n
x  u n  x  nu n1du  dx 
II. Tabular Integration
 f (x)g(x) dx , in which f can be differentiated repeatedly
Integrals of the form
to become zero and g can be integrated repeatedly without difficulty, can be
evaluated using tabular integration.
Example: Find
xe
3 2x
dx .
f (x ) and its derivatives
g (x ) and its antiderivatives
x3
(+)
e2x
3x 2
(–)
1 2x
e
2
6x
(+)
1 2x
e
4
6
(–)
1 2x
e
8
1 2x
e
16
0

1 3 2x 3 2 2x 6 2x 6 2x
x e  x e  xe  e  C
2
4
8
16
1
3
3
3
= x 3 e 2 x  x 2 e 2 x  xe2 x  e 2 x  C
2
4
4
8
x 3 e 2 x dx = 
III. Reduction Formulas
Integration by parts can be used to derive reduction formulas for integrals. These
are formulas that express an integral involving a power of a function in terms of an
integral that involves a lower power of that function
5
Example 1: Prove the reduction formula
use the result to find

 (ln x) dx  x(ln x)  n (ln x)
n
n
n 1
dx and
(ln x) 2 dx .

1 
Let u  (ln x) n and dv  dx  du  n(ln x) n 1  dx  and v  1dx  x . Thus,
x 
1 
(ln x) n dx  udv  uv  vdu  x(ln x) n  ( x)n(ln x) n 1  dx   x(ln x) n 
x 




n (ln x) dx . Thus, (ln x) dx  x(ln x)  2 ln x dx  x(ln x)



1dx]  x(ln x)  2x ln x  2x  C and this checks.
n 1
2
2
2
 2[ x(ln x) 
2
Example 2: Prove the reduction formula

n 1
(sin x) n2 dx .
n

1
(sin x) n dx   (sin x) n1 cos x 
n
Let u  (sin x) n1 and dv  sin x dx  du  (n  1)(sin x) n2 cos x dx and v =

sin x dx   cos x . Thus,


(sin x) n dx 


(sin x) n1 (sin x dx) 
 udv 
uv  vdu  (sin x) n1 ( cos x)  ( cos x)( n  1)(sin x) n2 cos x dx 

 (sin x) n1 cos x  (n  1) (sin x) n2 cos 2 x dx  (sin x) n1 cos x 


(n  1) (sin x) dx  n (sin x) dx  (sin x) cos x  (n  1) (sin x)



1
n 1
(sin x) dx   (sin x) cos x 
(sin x) dx .

n
n 
(n  1) (sin x) n2 (1  sin 2 x) dx  (sin x) n1 cos x  (n  1) (sin x) n2 dx 
n
n
n 1
n
n 1
n2
6
n2
dx 
Practice Sheet for Integration by Parts
(1)

3 x ln x dx =
(2) Prove the following reduction formula:


n
x m (ln x) n1 dx for m  1 .
m 1
(3)

(4)
 cos x  dx =
(5)

(6)

x arcsec x dx =
(7)

e x cos x dx =
(8)

x 3 sin x 2 dx =
(9)

e x dx =
(10)

x 2 e 3 x dx =
(11)

x 2 sin x dx =
arcsin x dx =
ln x
x3
dx =
 
7
x m (ln x) n dx 
1
x m1 (ln x) n 
m 1
(12)

(13)

(14)

3
e x dx 
3
e x
3x
dx 
sin(ln x)dx 
Solution Key for Integrations by Parts
(1) Let u  ln x and dv  x

1
3

1
3 4
1
dx and v  x 3 dx  x 3 
x
4
1
3 4
 3 4 3  1  3 4 3 3
x 3 dx  x 3 
 x  dx   x 
4
4
4
 x  4
dx  du 
3 x ln x dx = 3 x 4 3 ln x 
4


9 43
x C.
16

1 
(2) Let u  (ln x) n and dv  x m dx  du  n(ln x) n 1  dx  and v  x m dx 
x 
1
1
 1  1

n
x m 1  x m ln x  dx 
x m 1 (ln x) n  n(ln x) n 1  dx  
x m 1  
m 1
m x
 x  m 1




1
n
x m1 (ln x) n 
x m (ln x) n1 dx .
m x
m 1

arcsin x dx = x arcsin x 

1 x2
x
1 x

1
dx  x arcsin x 
2
1
(3) Let u  arcsin x and dv  dx  du 
2
x arcsin x  1  x 2  C.
8
dx and v  1dx  x 
 2x
1 x2
dx 
 cos x  dx = 2 wcos w dw . Let
u  w and dv  cos w dw  du  dw and v  cos w dw  sin w  cos x  dx =


2 wcos w dw = 2w(sin w)  2 (sin w) dw  2w sin w  2 sin w dw 



(4) Let w  x  w 2  x  2wdw  dx 
2w sin w  2 cos w  C  2 x sin( x )  2 cos( x )  C .

1
1
1
dx  du  dx and v  x 3 dx   x 2 
3
x
x
2
ln x
1
1  1  2
1
1
dx =  x  2 ln x 
x dx   x  2 ln x 
x 3 dx 


3
2
2
x
2
2
 
x
(5) Let u  ln x and dv 



1 2
1
 ln x
1
x ln x  x  2  C 
 2 C.
2
2
4
2x
4x
(6) Let u  arc sec x and dv  x dx  du 


x arcsec x dx =
1 2
1
x arc sec x 
2
2

1
dx and v 

x dx 
x x2 1
x
1
1 2
dx  x 2 arc sec x 
x 1  C .
2
2
2
x 1
(7) Let u  cos x and dv  e x dx  du   sin x dx and v 


e x dx  e x 
e x cos x  e x sin xdx . Use integration by parts to evaluate




e x cos x  e x sin x 




e x cos x dx 
e x sin x dx . Let
u  sin x and dv  e x dx  du  cos x dx and v  e x dx  e x 
e x sin x  e x cos x dx . Thus,
1 2
x 
2

e x sin x dx =
 e cos x dx = e cos x   e sin xdx =
x
x
x

e x cos x dx   e x cos x  e x sin x 

2 e x cos x dx  e x cos x  e x sin x 

e x cos x dx 
9
 e cos x dx 
x
ex
 cos x  sin x   C .
2
(8) Let w  x 2  dw  2 x dx 

 
x 3 sin x 2 dx 
1
2
 x sin  x 2x dx 
2
2

1
wsin w dw . Let u  w and dv  sin w dw  du  dw and v 
2
 cos w 
1
2

w sin w dw 
1
 w cos w 
2
 

 sin w dw 
1

cos w dw   w cos w 
2

 
1
1
1
sin w  C   x 2 cos x 2  sin x 2  C .
2
2
2
(9) Let w  x  w 2  x  2w dw  dx 
dv  e w dw  du  dw and v 
we w  e w 



e x dx = 2 we w dw . Let u  w and
 e dw  e   we dw  we   e dw 
w
w
w

w
e x dx = 2 we w dw  2we w  2e w  C  2 x e
w
x
 2e
(10) Use tabular integration:
f (x) and its derivatives
g (x) and its antiderivatives
x2
e 3x
(+)
1 3x
e
3
2x
(–)
1 3x
e
9
2
(+)
0
1 3x
e
27
Thus,

x 2 e 3 x dx =
1 2 3x 2 3x 2 3x
x e  xe 
e C.
3
9
27
10
x
C.
(11) Use tabular integration:
f (x) and its derivatives
g (x) and its antiderivatives
x2
sin x
(+)
 cos x
2x
(–)
 sin x
2
(+)
cos x
0
Thus,

x 2 sin x dx =  x 2 cos x  2 x sin x  2 cos x  C .
(12) Let w  3 x  w3  x  3w 2 dw  dx 


 w e dw can
3
e x dx  3 w 2 e w dw .
2 w
integrated using tabular integration:
f (x) and its derivatives
g (x) and its antiderivatives
w2
(+)
2w
(–)
2
(+)
ew
ew
ew
ew
0

w 2 e w dw = w 2 e w  2we w  2e w  C 
33 x2 e
3
x
6 3 xe
3
x
3

3
 6e x + C.
3
(13) Let w  x  w  x  3w dw  dx 
3

e x dx  3 w 2 e w dw 
2
In solution to problem #9 previously,


e
3
3
x


w2e w
dw  3 we w dw .
dx  3
w
x
we dw  we  e 
11
w
w
w

e
3
3
x
x
dx 

3 we w dw  3we w  3e w  C  3 3 x e
3
x
 3e
3
x
C .

1 
(14) Let u  sin(ln x) and dv  dx  du  cos(ln x) dx  and v  1dx  x 
x 

sin(ln x)dx  x sin(ln x)  cos(ln x) dx . Use integration by parts to evaluate

1 
cos(ln x) dx . Let u  cos(ln x) and dv  dx  du   sin(ln x) dx  and
x 


v  1dx  x 
 cos(ln x) dx = x cos(ln x)   sin(ln x) dx . Thus,  sin(ln x)dx 




x sin(ln x)  cos(ln x) dx = x sin(ln x)   x cos(ln x)  sin(ln x) dx   x sin(ln x) 




x cos(ln x)  sin(ln x) dx  2 sin(ln x) dx  x sin(ln x)  x cos(ln x) 

sin(ln x)dx 
1
1
x sin(ln x)  x cos(ln x)  C .
2
2
12