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Math Help - Can someone explain this to me?

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

    The given is:

    \int\frac{tan^2 e^{-x}}{e^x}dx<br />

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

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


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


    If you rewrite the original given it would be,

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