Can someone explain this to me?

**honestliar** · August 29th 2009, 06:59 AM

The given is:

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

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

**Plato** · August 29th 2009, 07:11 AM

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

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