Calculus 6.3 day 1

Calculus 6.3 day 1

6.3 day 1 Integration By Parts Badlands, South Dakota Photo by Vickie Kelly, 1993 Greg Kelly, Hanford High School, Richland, Washington 6.3 day 1 Integration By Parts Start with the product rule: d du dv uv v u dx dx dx d uv du v 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 zero (usually). dv is easy to integrate. The Integration by Parts formula is a product rule for integration. Choose u in this order:

LIPET Logs, Inverse trig, Polynomial, Exponential, Trig Example: x cos x dx polynomial factor u v v du u dv uv v du LIPET u x dv cos x dx du dx v sin x x sin x sin x dx x sin x cos x C Another Example: 2 x x e dx u v v du x e e 2 x dx 2 x u dv uv v du u x 2 du 2 x dx

x 2 x x x e 2xe dx 2 x x x x x x e 2 xe e dx 2 x LIPET dv e x dx v e x This is still a product, so we x u x integration need to use by

dv e dx parts again. du dx v e x x e 2 xe 2e C Still Another: LIPET x e cos x dx u v v du e x sin x sin x e x dx x u e x dv cos x dx du e x dx v sin x x dv sin x dx u e x du e dx v cos x x x e sin x e cos x cos x e dx

uv v du e x sin x e x cos x e x cos x dx This is the expression we started with! One More: LIPET x e cos x dx u v v du e x sin x sin x e x dx x x u e x dv cos x dx du e x dx v sin x x dv sin x dx u e x du e dx v cos x x

e sin x e cos x cos x e dx x x x x e cos x dx e sin x e cos x e cos x dx 2 e x cos x dx e x sin x e x cos x x x e sin x e cos x x C e cos x dx 2

One More: e x cos x dx u v v du e x sin x sin x e x dx x x This is called solving for the unknown integral. It works when both factors integrate and differentiate forever. x e sin x e cos x cos x e dx x x x x e cos x dx e sin

x e cos x e cos x dx 2 e x cos x dx e x sin x e x cos x x x e sin x e cos x x C e cos x dx 2 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 x x e dx f x & deriv. g x & integrals 2 e 2x ex x 2 0 x e x e x Compare this with the same problem done the other way: 2 x x x x e

2 xe 2 e C x e dx 2 x 2 x x e dx u v v du x e e 2 x dx 2 x u dv uv v du u x 2 du 2 x dx x x 2 e x 2xe x dx 2 x x u x x x e 2 xe e dx 2 x x

x x e 2 xe 2e C du dx LIPET dv e x dx v e x x dv e dx v e x This is easier and quicker to do with tabular integration! Another example: 3 x sin x dx 3 x sin x 3x 2

cos x 6x 6 sin x 0 sin x cos x x 3 cos x 3 x 2 sin x 6 x cos x 6sin x + C Now lets try tabular integration for a solving for the unknown integral problem: x e cos x dx x cos x ex sin x ex cos x e We recognize the

integrand in this line! We could write the problem like this: x x x e sin x e cos x e cos x dx e cos x dx x 2 e x cos x dx e x sin x e x cos x x x e sin x e cos x x C e cos x dx 2

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