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Math Help - Simple Integration problem: Trigonometric Integration

  1. #1
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    Simple Integration problem: Trigonometric Integration

    I'm not sure what I did wrong here, or if the integral calculator has given me an equivalent answer, I'll try my best to explain.

    Question: find the indefinite integral of cos(x)*cos^5(sin(x))dx

    First step is easy, to sub u=sin(x), du=cos(x)dx

    = int cos^5(u)du

    Then I subbed the Cos^2(x)=1-Sin^2(x) formula into the equation to get

    = int (1-sin^2(u))^2*cos(u)du then sub v=sin(u), dv=cos(u)du
    = int (1-v^2)^2*dv
    = int v^4 - 2v^2 + 1
    = 1/5*v^5 - 2/3*v^3 + v + C
    = 1/5*sin^5(u) - 2/3*sin^3(u) + sin(u) + C
    = 1/5*sin^5(sin(x)) - 2/3*sin^3(sin(x)) + sin(sin(x)) + C

    An integral calculator gives an answer of 1/80*(sin(5*sin(x)) + 5/48(sin(3*sin(x)) + 5/8*(sin(sin(x))

    Thats alot different from my answer, what did I do wrong?
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  2. #2
    Ted
    Ted is offline
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    Hello.
    Am sure, your solution is correct.
    Maybe, the calculator's solution is a different form for your answer as you said.
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  3. #3
    MHF Contributor
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    Hi

    You did nothing wrong since

    1/5*sin^5(sin(x)) - 2/3*sin^3(sin(x)) + sin(sin(x)) = 1/80*(sin(5*sin(x)) + 5/48(sin(3*sin(x)) + 5/8*(sin(sin(x))
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