Question
Evaluate the Integral
pi
/ = (3cos^2x-1) (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!
pi
/ = (3cos^2x-1) (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!
Answers
Steve
∫[0,pi] (3cos^2x-1) (sinx) dx
let u = cosx, then du = -sinx dx
now you have
∫[1,-1] -(3u^2-1) du
= u^3 - u [-1,1]
= 0
let u = cosx, then du = -sinx dx
now you have
∫[1,-1] -(3u^2-1) du
= u^3 - u [-1,1]
= 0