Math
posted by Rianne .
Evaluate the Integral
pi
/ = (3cos^2x1) (sinx) dx=0
0
I am really confused with this, please help!
** used / for integral, not sure if there is a way to type it!

∫[0,pi] (3cos^2x1) (sinx) dx
let u = cosx, then du = sinx dx
now you have
∫[1,1] (3u^21) du
= u^3  u [1,1]
= 0