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8.2 Integration By Parts
Badlands, South DakotaGreg Kelly, Hanford High School, Richland, WashingtonPhoto by Vickie Kelly, 1993
cos x x dx
Integrate the following:
8.2 Integration By Parts
Start with the product rule:
d du dvuv v u
dx dx 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
The Integration by Parts formula is a “product rule” for integration.
u differentiates to zero (usually).
dv is easy to integrate.
Choose u in this order: LIPET
Logs, Inverse trig, Polynomial, Exponential, Trig
Or LIPTE
Integration by Parts
!
Example 1:
cos x x dxpolynomial factor u x
du dx
cos dv x dx
sinv x
u dv uv v du LIPET
sin cosx x x C
u v v du
sin sin x x x dx
cos x x dx
Example 2:
ln x dxlogarithmic factor lnu x
1du dx
x
dv dx
v x
u dv uv v du LIPET
lnx x x C
1ln x x x dx
x
u v v du
ln x dx
This is still a product, so we need to use integration by parts again.
Example 3:2 xx e dx
u dv uv v du LIPET
2u x xdv e dx
2 du x dx xv e u v v du
2 2 x xx e e x dx 2 2 x xx e xe dx u x xdv e dx
du dx xv e 2 2x x xx e xe e dx 2 2 2x x xx e xe e C
2 xx e dx
Example 4:
cos xe x dxLIPET
xu e sin dv x dx xdu e dx cosv x
u v v du sin sinx xe x x e dx
sin cos cos x x xe x e x x e dx
xu e cos dv x dx xdu e dx sinv x
sin cos cos x x xe x e x e x dx This is the expression we started with!
uv v du
Example 4 (con’t):
cos xe x dx u v v du
cos xe x dx 2 cos sin cosx x xe x dx e x e x
sin coscos
2
x xx e x e x
e x dx C
sin sinx xe x x e dx xu e sin dv x dx
xdu e dx cosv x
xu e cos dv x dx xdu e dx sinv x
sin cos cos x x xe x e x e x dx
sin cos cos x x xe x e x x e dx
Example 4 (con’t):
cos xe x dx u v v du
This is called “solving for the unknown integral.”
It works when both factors integrate and differentiate forever.
cos xe x dx 2 cos sin cosx x xe x dx e x e x
sin coscos
2
x xx e x e x
e x dx C
sin sinx xe x x e dx
sin cos cos x x xe x e x e x dx
sin cos cos x x xe x e x x e dx
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:
2 xx e dx & deriv.f x & integralsg x
2x
2x
2
0
xexexexe
2 xx e dx 2 xx e 2 xxe 2 xe C
Compare this with the same problem done the other way:
Example 5:2 xx e dx
u dv uv v du LIPET
2u x xdv e dx
2 du x dx xv e u v v du
2 2 x xx e e x dx 2 2 x xx e xe dx u x xdv e dx
du dx xv e 2 2x x xx e xe e dx
2 2 2x x xx e xe e C This is easier and quicker to do with tabular integration!
3 sin x x dx3x
23x
6x
6
sin x
cos x
sin xcos x
0
sin x
3 cosx x 2 3 sinx x 6 cosx x 6sin x + C
Find
You Try:
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