Integration By Parts

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Sharp Tutor Wed, Jan 19, 2022 07:17 PM UTC

This presentation discusses integration by parts with some relevant examples.

1. 6.3 Integration By Parts
Badlands, South Dakota
Photo by Vickie Kelly, 1993 Greg Kelly, Hanford High School, Richland, Washington

2. 6.3 Integration By Parts
Start with the product rule:
u dv  d  uv   v du 
d dv du
 uv  u  v
dx dx dx
u dv  d  uv    v du
d  uv  u dv  v du
d  uv   v du u dv u dv uv  v du
This is the Integration by Parts
u dv d  uv   v du formula.


3. u dv uv  v du
dv is easy to
u differentiates 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


4. Example 1:
u dv uv  v du
x cos x dx LIPET
polynomial factor u x dv cos x dx
du dx v sin x
u v  v du
x sin x  sin x dx
x sin x  cos x  C


5. u dv uv  v du
ln x dx LIPET
logarithmic factor u ln x dv dx
u v  v du 1
du  dx v x
x
1
ln x x  x  dx
x
x ln x  x  C


6. Example 4:
2 x u dv uv  v du LIPET
x e dx
u x 2 dv e x dx
u v  v du
du 2 x dx v e x
2 x x
x e  e 2 x dx
2 x x This is still a product, so we
x e  2xe dx u
need to use x integration
dv e dxx
by
parts again.
du dx v e x
2 x

x e  2 xe  e dx x x

2 x x x
x e  2 xe  2e  C


7. Example 5: LIPET
x u e x dv cos x dx
 cos x dx
e
du e x dx v sin x
u v  v du x
u e dv sin x dx
x x
e sin x  sin x e dx x
du e dx v  cos x
x
 x
e sin x  e  cos x   cos x e dx x

uv v du This is the
x x x
e sin x  e cos x  e cos x dx expression we
started with!


8. Example 6: LIPET
x u e x dv cos x dx
 cos x dx
e
du e x dx v sin x
u v  v du x
u e dv sin x dx
x x
e sin x  sin x e dx x
du e dx v  cos x
x
 x
e sin x  e  cos x   cos x e dx x

x x x x
e cos x dx e sin x  e cos x  e cos x dx
x x x
2 e cos x dx e sin x  e cos x
x x
x e sin x  e cos x
e cos x dx  2
C

9. Example 6:
x
This is called “solving for
e cos x dx the unknown integral.”
u v  v du It works when both
x x factors integrate and
e sin x  sin x e dx differentiate forever.
x
 x
e sin x  e  cos x   cos x e dx x

x x x x
e cos x dx e sin x  e cos x  e cos x dx
x x x
2 e cos x dx e sin x  e cos x
x x
x e sin x  e cos x
e cos x dx  2
C


10. A Shortcut: Tabular Integration
Tabular integration works for integrals of the form:
f  x  g  x  dx
where: Differentiates to Integrates
zero in several repeatedly.
steps.


11. 2 x
 e dx
x
f  x  & deriv. g  x  & integrals
2 x
 x e
 2x ex
Compare this with
 2 e x the same problem
done the other way:
x
0 e
2 x x x
2 x
x e dx  x e  2 xe 2e C


12. Example 5:
2 x u dv uv  v du LIPET
x e dx
u x 2 dv e x dx
u v  v du
du 2 x dx v e x
2 x x
x e  e 2 x dx
x 2 e x  2xe x dx u x dv e dxx
du dx v e x
2 x

x e  2 xe  e dx x x

This is easier and quicker to
2 x x x
x e  2 xe  2e  C do with tabular integration!


13. 3
 sin x dx
x
3
 x sin x
 3x 2  cos x
 6x  sin x
 6 cos x
0 sin x
 x 3 cos x  3 x 2 sin x  6 x cos x  6sin x + C


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