# Thread: What is wrong with my solution?

1. ## What is wrong with my solution?

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

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

$u=e^{-x}$
$-du=e^{-x}$

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

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

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

what is wrong with my solution?

the answer should be:
$-tan e^{-x} + e^{-x} + c$

Thanks

2. your integral is

-int(sec^(u) -1 )du

=- tan(u) - u + c

= -tan(e^(-x)) + e^(-x) +c