Posted by **Rianne** on Saturday, November 10, 2012 at 10:52pm.

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!

- Math -
**Steve**, Sunday, November 11, 2012 at 1:00am
∫[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

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