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Math Help - What is wrong with my solution?

  1. #1
    Newbie
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    What is wrong with my solution?

    \int \frac{tan^2 e^{-x}}{e^x}dx

    \int tan^2 e^{-x}e^{-x}dx

    u=e^{-x}
    -du=e^{-x}


     - \int sec^{2}-1 udu

    - \int sec^{2}dx -\int dx udu

     -tan-x+ e^{-x}+c


    what is wrong with my solution?

    the answer should be:
    -tan e^{-x} + e^{-x} + c


    Thanks
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  2. #2
    MHF Contributor Calculus26's Avatar
    Joined
    Mar 2009
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    Florida
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    1,271
    your integral is

    -int(sec^(u) -1 )du

    =- tan(u) - u + c

    = -tan(e^(-x)) + e^(-x) +c
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