6.3 Integration By Part Badlands, South Dakota Greg Kelly, Hanford High School, Richland, Washington Photo by Vickie Kelly, 1993
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6.3 Integration By Parts
Badlands, South DakotaGreg Kelly, Hanford High School, Richland, WashingtonPhoto by Vickie Kelly, 1993
6.3 Integration By Parts
Start with the product rule:
d dv duuv u vdx 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
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
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
Example:
ln x dxlogarithmic factor lnu x
1du dxx
dv dx
v x
u dv uv v du LIPET
lnx x x C
1ln x x x dxx
u v v du
This is still a product, so we need to use integration by parts again.
Example 4:2 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
Example 5:
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 6:
cos xe x dxLIPET
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 xe 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 6:
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 xe 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
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 xx e dx & deriv.f x & integralsg x
2x
2x
20
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 dx3x23x
6x
6
sin xcos x
sin xcos x
0
sin x
3 cosx x 2 3 sinx x 6 cosx x 6sin x + C
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