Posted by Michael on Monday, December 1, 2008 at 3:54pm.
y' = 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
let 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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