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December 21, 2014

December 21, 2014

Posted by **Michael** on Monday, December 1, 2008 at 3:54pm.

- calc -
**Reiny**, Monday, December 1, 2008 at 4:16pmy' = 20 * (sinx)^3 * (cosx)^2

= 20 sinx(sinx)^2(cosx)^2

= 20sinx(1 - cos^2 x)cos^2 x

= 20sinx cos^2 x - 20sinx cos^4 x

y = -20/3(cos^3 x) + 20/3(cos^5 x) + C, where C is a constant

- calc -
**Damon**, Monday, December 1, 2008 at 4:21pmlet u = sin^2 x

then du = 2 sin x cos x dx

let dv = -3 sin x cos^2 x dx

then v = cos^3 x

u dv = -3 sin^3 x cos^2 x dx

so

let u = -20/3 sin^2 x

then du = -40/3 sin x cos x dx

let dv = -(1/3)sin x cos^2 x dx

then v = cos^3 x

now by parts

int u dv = u v - int v du

int (20 sin^3 x cos^2 x dx) =

-(20/3)sin^2 x cos^3 x +(40/3)int(cos^4 x sin x dx)

well remember the first term at the end and the (40/3) and work on the integral

int cos^4 x sin x dx = -(1/5)cos^5 x

I think you can get it from there

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