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Math Help - integration by parts

  1. #1
    HyrdaD
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    integration by parts

    Use integration by parts to deduce the following formulas.

    A) int_ sin^3[x] dx = (-3/4)cos[x] + (1/12)cos[3x]

    B) int_ sin^4[x] dx = (3/8)x -(1/4)sin[2x] + (1/32)sin[4x]

    Thanks so much
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  2. #2
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    Quote Originally Posted by HyrdaD View Post
    A) int_ sin^3[x] dx = (-3/4)cos[x] + (1/12)cos[3x]
    Note,
    sin^3 x = sin^2 x * sin x = (1 - cos^2 x)*sin x

    That means use the substitution t = cos x.

    B) int_ sin^4[x] dx = (3/8)x -(1/4)sin[2x] + (1/32)sin[4x]
    Note,
    sin^4 x = (sin^2 x)^2 = (1/2 - 1/2*cos 2x)^2
    Expand,
    1/4 - 1/2*cos 2x + 1/4*cos^2 (2x)
    You can use another half-angle formula on cosine.
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