Transcript Calc06_3 integration by parts

6.3 Integration By Parts
Badlands, South Dakota
Photo by Vickie Kelly, 1993
Greg Kelly, Hanford High School, Richland, Washington
6.3 Integration By Parts
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.

 u dv  uv   v du
u differentiates to
dv is easy to
integrate.
zero (usually).
The Integration by Parts formula is a “product rule” for
integration.
Choose u in this order:
LIPET
Logs, Inverse trig, Polynomial, Exponential, Trig

Example 1:
 x  cos x dx
polynomial factor
u v   v du
 u dv  uv   v du
LIPET
ux
dv  cos x dx
du  dx
v  sin x
x  sin x   sin x dx
x  sin x  cos x  C

Example 4:
 ln x dx
logarithmic factor
u v   v du
 u dv  uv   v du
LIPET
u  ln x
dv  dx
1
du  dx
x
vx
1
ln x  x   x  dx
x
x ln x  x  C

Example 5:
 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

This is still a product, so we
x
u
x integration
need to
use
by
dv  e dx
parts again.
du  dx
v  ex
x 2 e x  2 xe x  2e x  C

Example 6:
e
x
LIPET
u  cos x
cos x dx
u v   v du
cos x  e   e   sin x dx
x
x
cos x  e x   e x sin x dx
du   sin x dx
ve
u  sin x
dv  e dx
x
x
x
x
du  cos x dx
cos x  e  sin x  e   e cos x dx
x
dv  e x dx
v  ex
This is the
expression we
started with!

Example 6:
e
x
LIPET
u  cos x
cos x dx
u v   v du
cos x  e   e   sin x dx
x
x
cos x  e x   e x sin x dx
du   sin x dx
ve
u  sin x
dv  e dx
x
x
x
x
x
x
sin
x

e

cos
x

e
x
C
 e cos x dx 
2
x
x
du  cos x dx
 e cos x dx  cos x  e  sin x  e   e
2  e cos x dx  cos x  e  sin x  e
x
x
dv  e x dx
x
v  ex
cos x dx
Example 6:
e
x
This is called “solving
for the unknown
integral.”
cos x dx
u v   v du
cos x  e   e   sin x dx
x
x
cos x  e x   e x sin x dx
It works when both
factors integrate and
differentiate forever.
 e cos x dx  cos x  e  sin x  e   e
2  e cos x dx  cos x  e  sin x  e
x
x
x
x
x
x
cos x dx
x
x
x
sin
x

e

cos
x

e
x
C
 e cos x dx 
2

A Shortcut: Tabular Integration
Tabular integration works for integrals of the form:
 f  x  g  x  dx
where:
Differentiates to
zero in several
steps.
Integrates
repeatedly.

2 x
x
 e dx
f  x  & deriv. g  x  & integrals
 x
2
e
 2x
ex
 2
0
x
e
x
e
x
Compare this with
the same problem
done the other way:
x
2 x
x

2
e
x
e

2
xe
C
 x e dx 
2 x

Example 5:
 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

ux
x e  2 xe   e dx
2 x
LIPET
x
x

x 2 e x  2 xe x  2e x  C
du  dx
dv  e dx
x
v  ex
This is easier and quicker to
do with tabular integration!

x
3
x

3
sin x dx
 3x 2
sin x
 cos x
 6x
 6
 sin x
cos x
0
sin x
 x 3 cos x  3x 2 sin x  6x cos x  6sin x + C
p