integration by parts. integration by parts start with the product rule: this is the integration by...
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![Page 1: Integration by Parts. Integration By Parts Start with the product rule: This is the Integration by Parts formula](https://reader035.vdocuments.site/reader035/viewer/2022062318/551bf5215503469e4f8b4614/html5/thumbnails/1.jpg)
Integration by PartsIntegration by Parts
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Integration By Parts
Start with the product rule:
d dv duuv 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
This is the Integration by Parts formula.
dxvuuvdxvu
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The Integration by Parts formula is a “product rule” for integration.
u differentiates to zero (usually).
v’ is easy to integrate.
Choose u in this order: LIPET
Logs, Inverse trig, Polynomial, Exponential, Trig
dxvuuvdxvu
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Example 1:
polynomial factor u x
sinv x
LIPET
sin sin x x x dx
dxxx cos
xv cos
dxvuuvdxvu
1u dxvuuvdxvu
cxxxdxxx cossincos
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Example 2:
logarithmic factor
LIPET
dxxln
1v
dxvuuvdxvu
xu
1
dxvuuvdxvu
cxxxdxx ln1ln
xu ln
xv
1ln x x x dx
x
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This is still a product, so we need to use integration by parts again.
Example 3:2 xx e dx
LIPET
2u xxv e
u xxv e
dxvuuvdxvu
xu 2
xev
1u
xev
dxvuuv
dxxeex xx 22
dxxeex xx 22
dxexeex xxx 22
cexeexdxex xxxx 2222
![Page 7: Integration by Parts. Integration By Parts Start with the product rule: This is the Integration by Parts formula](https://reader035.vdocuments.site/reader035/viewer/2022062318/551bf5215503469e4f8b4614/html5/thumbnails/7.jpg)
dxvuuvdxvu
u dv uv v du
The Integration by Parts formula can be written as:
![Page 8: Integration by Parts. Integration By Parts Start with the product rule: This is the Integration by Parts formula](https://reader035.vdocuments.site/reader035/viewer/2022062318/551bf5215503469e4f8b4614/html5/thumbnails/8.jpg)
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
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Example 4 continued …
cos xe x dx LIPET
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
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Example 4 continued …
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
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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.
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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:
![Page 13: Integration by Parts. Integration By Parts Start with the product rule: This is the Integration by Parts formula](https://reader035.vdocuments.site/reader035/viewer/2022062318/551bf5215503469e4f8b4614/html5/thumbnails/13.jpg)
This is still a product, so we need to use integration by parts again.
Example 3:2 xx e dx
LIPET
2u xxv e
u xxv e
dxvuuvdxvu
xu 2
xev
1u
xev
dxvuuv
dxxeex xx 22
dxxeex xx 22
dxexeex xxx 22
cexeexdxex xxxx 2222
This is easier and quicker to do with tabular integration!
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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
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Acknowledgement
I wish to thank Greg Kelly from Hanford High School, Richland, USA for his hard work in creating this PowerPoint.
http://online.math.uh.edu/
Greg has kindly given permission for this resource to be downloaded from www.mathxtc.com and for it to be modified to suit the Western Australian Mathematics Curriculum.
Stephen CorcoranHead of MathematicsSt Stephen’s School – Carramarwww.ststephens.wa.edu.au