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Math

posted by on .

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 - ,

    ∫[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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