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Math Help - Integration discrepancy with wolfram Online

  1. #1
    Newbie
    Joined
    Aug 2010
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    21

    Integration discrepancy with wolfram Online

    Integrate:

    (4+sin(x))^3(cos(x))dx

    This is what I did:
    Let u=sin(x)+4

    \frac{du}{dx}=cos(x)

    du=cos(x)dx
    Therefore, I have:

    \int_ (u^3)du

    Now,I get

    \frac{u^4}{4}+c

    Subbing back in, I get

    \frac{1}{4}(sin(x)+4)^4 which when expanded out gives me

    \frac{1}{4}(sin^4(x)+16sin^3(x)+96sin^2(x)+256sin(  x)+256)

    Then, quartering everything:

    \frac{sin^4(x)}{4}+4sin^3(x)+24sin^2(x)+64sin(x)+6  4

    The problems is that Wolfram Online Integrator does not include the final term "64" in the answer.

    Where have I gone wrong??
    Thanks in advance
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  2. #2
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    64 is a constant, so the answer remains the same whether you have the 64 or you don't have it. Wolfram's answer is different than yours in that they have the constant of integration c' = c + 64 where your constant is c.
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