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

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

$\displaystyle u=e^{-x}$

$\displaystyle -du=e^{-x}$

$\displaystyle - \int sec^{2}-1 udu$

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

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

what is wrong with my solution?

the answer should be:

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

Thanks