6.1 Integration by parts
Formula for Integration by parts
 f ( x) g '( x)dx  f ( x) g ( x)   g ( x) f '( x)dx
 udv  uv   vdu
• The idea is to use the above formula to simplify an integration task.
• One wants to find a representation for the function to be integrated
in the form udv so that the function vdu is easier to integrate
than the original function.
• The rule is proved using the Product Rule for differentiation.
Deriving the Formula
Start with the product rule:
d
dv
du
 uv   u  v
dx
dx
dx
d  uv   u dv  v du
d  uv   v du  u dv
u dv  d  uv   v du
 u dv    d  uv   v du 
 u dv    d  uv     v du
 u dv  uv   v du
This is the Integration by Parts
formula.
Choosing u and v
 u dv  uv   v du
u differentiates to
dv is easy to
zero (usually).
integrate.
Choose u in this order:
LIPET
Logs, Inverse trig, Polynomial, Exponential, Trig

Example 1:
 x  cos x dx
polynomial factor
u v   v du
x  sin x   sin x dx
x  sin x  cos x  C
 u dv  uv   v du
LIPET
ux
dv  cos x dx
du  dx
v  sin x
Example 2:
 ln x dx
logarithmic factor
u v   v du
1
ln x  x   x  dx
x
x ln x  x  C
 u dv  uv   v du
LIPET
u  ln x
dv  dx
1
du  dx
x
vx
Example 3:
 x e dx
u v   v du
x e   e  2 x dx
2 x
2 x
 u dv  uv   v du
u  x2
dv  e x dx
du  2 x dx
ve
x
x
x e  2  xe dx
2 x
x

x e  2 xe   e dx
2 x
LIPET
x
x

x 2 e x  2 xe x  2e x  C
This is still a product, so we
x
u
x integration
need to
use
by
dv  e dx
parts again.
du  dx
v  ex
Example 4:
LIPET
u  e x dv  cos x dx
du  e x dx v  sin x
x
e
 cos x dx
u v   v du
ue
dv  sin x dx
x
du  e dx v   cos x
x
e sin x   sin x  e dx
x
x

e sin x  e   cos x    cos x  e dx
x
x
x
uv
v du
e sin x  e cos x   e cos x dx
x
x
x

This is the
expression we
started with!

Example 4(cont.):
LIPET
u  e x dv  cos x dx
du  e x dx v  sin x
x
e
 cos x dx
u v   v du
ue
dv  sin x dx
x
du  e dx v   cos x
x
e sin x   sin x  e dx
x
x

e sin x  e   cos x    cos x  e dx
x
x
x

 e cos x dx  e sin x  e cos x   e
2 e cos x dx  e sin x  e cos x
x
x
x
x
x
x
x
x
e
sin
x

e
cos x
x
C
 e cos x dx 
2
x
cos x dx
This is called “solving for
the unknown integral.”
It works when both factors
integrate and differentiate
forever.
Integration by Parts
for Definite Integrals
b
b
b
a
a
 udv  uv    vdu  u  b  v  b   u a v a    vdu
Formula
b
a
a
Integration by Parts Formula and the Fundamental Theorem of Calculus
imply the above Integration by Parts Formula for Definite Integrals. Here
we must assume that the functions u and v and their derivatives are
all continuous.
Example
1
2
 arcsin  x  dx
0
1
2
Choose u  arcsin  x  . Then
dv  dx, v  x and du 
1
2
 arcsin  x  dx  x arcsin  x 0  
0
1
2
0
dx
1 x
xdx
1 x 2
2
.
Integration by Parts
for Definite Integrals
Example (cont’d)
By the computations on the previous slide
we now have
1
2
1
2
 arcsin  x  dx  x arcsin  x 0  
1
2
0
0
1
2
Compute

0
xdx
1 x 2
xdx

1 x 2
arcsin  12 
2
1
2

0
xdx
1 x 2
by the substitution t  1  x 2, dt  2 xdx.
x  0  t  1 and x 
1
3
t 
2
4
1
2
3
xdx

1 x 2
0

dt
3
   3   1  2  3


t
1 2 t

2
2
1
4
4
Combining these results we get the answer
1
2
 arcsin  x  dx 
0
arcsin  12 
2

2 3  2 3


2
12
2