calc
posted by Michael on .
What is the integral of 20 * (sinx)^3 * (cosx)^2?

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