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Thread: Can someone explain this to me?

  1. #1
    Newbie
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    Can someone explain this to me?

    The given is:

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

    and the answer is:
    $\displaystyle -tan e^{-x} + e^{-x} + c$

    MY answer is:
    $\displaystyle -tan e^{-x}+ c$ yeah, I know I'm wrong


    So, can someone explain how did my book got that second term($\displaystyle e^{-x}$)?


    If you rewrite the original given it would be,

    -$\displaystyle \int sec^{2}-1 u du$


    So where did it get the second term?


    thanks
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  2. #2
    MHF Contributor

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    $\displaystyle \begin{gathered}
    \int \frac{{\tan ^2 \left( {e^{ - x} } \right)}}
    {{e^x }}dx = \int {e^{ - x} \tan ^2 \left( {e^{ - x} } \right)dx} \hfill \\
    u = e^{ - x} \; \Rightarrow \;\int { - \tan ^2 (u)du = - \tan (u) + u} \hfill \\
    \end{gathered} $
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