math

what is the integral of sin^5(6x)dx

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asked by walter
  1. first, note that if u = 6x, you have

    1/6 ∫ sin^5 u du
    = 1/6 ∫sin^4(u) sinu du
    = -1/6 ∫(1-cos^2 u)^2 (-sinu du)
    If v = cos u, then dv = -sinu du, and you have

    -1/6 ∫(1-v^2)^2 dv

    just expand that and it's a regular old polynomial.

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    posted by Steve

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