# Math

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Evaluate the Integral

pi
/ = (3cos^2x-1) (sinx) dx=0
0

** used / for integral, not sure if there is a way to type it!

• Math -

∫[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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