# Can someone explain this to me?

• Aug 29th 2009, 07:59 AM
honestliar
Can someone explain this to me?
The given is:

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

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

$-tan e^{-x}+ c$ 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
• Aug 29th 2009, 08:11 AM
Plato
$\begin{gathered}
\int \frac{{\tan ^2 \left( {e^{ - x} } \right)}}
{{e^x }}dx = \int {e^{ - x} \tan ^2 \left( {e^{ - x} } \right)dx} \hfill \\
u = e^{ - x} \; \Rightarrow \;\int { - \tan ^2 (u)du = - \tan (u) + u} \hfill \\
\end{gathered}$