Transcript 8.2 Integration by Parts

8.2 Integration By Parts
Badlands, South Dakota
Photo by Vickie Kelly, 1993
Greg Kelly, Hanford High School, Richland, Washington
Integrate the following:
 x  cos x
dx
8.2 Integration By Parts
Start with the product rule:
d
dx
 uv   v
du
dx
u
dv
dx
d uv   v du  u dv
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
Or
LIPTE
Logs, Inverse trig, Polynomial, Exponential, Trig

Integration by Parts
!
Example 1:
 x  cos x
u
dx
polynomial factor
u v   v du
dv  uv   v du
LIPET
ux
d v  co s x d x
du  dx
v  sin x
x  sin x   sin x dx
 x  cos x
dx  x  sin x  co s x  C

Example 2:
u
 ln x dx
dv  uv   v du
LIPET
logarithmic factor
u v   v du
u  ln x
du 
1
dx
dv  dx
vx
x
ln x  x 
1
 x x
dx
x ln x  x  C
ln
x
dx



Example 3:
x
2
u
x
e dx
dv  uv   v du
u  x
u v   v du
du  2 x dx
x e   e  2 x dx
2
x
x
x

x e  2 xe 
2

dv  e dx
x
ve
x
x
x e  2  xe dx
2
2
LIPET
x
x
 e dx
x

This is still a product, so we
x
u
x integration
need to
use
by
dv  e dx
parts again.
x
du  dx
ve
x e dx  x 2 e x  2 xe x  2 e x  C
2
x

Example 4:
e
x
LIPET
u e
cos x dx
x
u e
e sin x   sin x  e dx
x
x

e sin x  e   cos x 
x
  cos x  e dx
x
v du
x
v  sin x
d v  sin x d x
du  e dx
e sin x  e cos x   e cos x dx
x
x
x
uv
x
d v  co s x d x
du  e dx
u v   v du
x
x
v   cos x

This is the
expression we
started with!

Example 4 (con’t):
e
x
u e
cos x dx
x
u e
e sin x   sin x  e dx
x
x

d v  co s x d x
du  e dx
u v   v du
e sin x  e   cos x 
x
x
x
x
d v  sin x d x
du  e dx
x
  cos x  e dx
x
v  sin x
v   cos x

 e cos x dx  e sin x  e cos x   e cos x dx
x
x
x
x
2  e cos x dx  e sin x  e cos x
x
x
x
e sin x  e cos x
x
 e cos x dx 
x
x
2
C
Example 4 (con’t):
e
x
This is called “solving
for the unknown
integral.”
cos x dx
u v   v du
It works when both
factors integrate and
differentiate forever.
e sin x   sin x  e dx
x
x

e sin x  e   cos x 
x
x
  cos x  e dx
x

 e cos x dx  e sin x  e cos x   e cos x dx
x
x
x
x
2  e cos x dx  e sin x  e cos x
x
x
x
e sin x  e cos x
x
 e cos x dx 
x
x
2
C

Integration by Parts
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.
Such as:


f
x

2
x
x e dx
g  x  & integrals
& d eriv.
x
2
e
 2x

2
0

x
e
x
e
x
e
x
Compare this with
the same problem
done the other way:
x e dx  x e  2 xe  2 e  C
2
x
2
x
x
x

Example 5:
x
2
u
x
e dx
dv  uv   v du
u  x
u v   v du
2
du  2 x dx
x e   e  2 x dx
2
x
x
x

x e  2 xe 
2
x
x
e
x
dx

x e  2 xe  2 e  C
2
dv  e dx
x
ve
x
x
x e  2  xe dx
2
LIPET
x
x
x
ux
dv  e dx
du  dx
ve
x
x
This is easier and quicker to
do with tabular integration!



3
x sin x dx
3
sin x
 3x2
 cos x
 6x
 sin x
 6
cos x
0
sin x
x
 x cos x  3 x sin x  6 x co s x  6 sin x + C
3
2
p
You Try:
Find
Solution:
Begin as usual by letting u = x2 and dv = v' dx = sin 4x dx. Next, create
a table consisting of three columns, as shown.
Homework:
Day 1: pg. 531, 11-55 EOO, 59-69 odd.
Day 2: MMM BC pgs. 106-107