# Thread: [SOLVED] how to start integrating e^(-x)*tan(e^(-x))dx

1. ## [SOLVED] how to start integrating e^(-x)*tan(e^(-x))dx

I don't know how to begin with this problem:

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

I've tried u substitution using $\displaystyle {u = {e^{ - x}}}$ as well as integrations by parts using $\displaystyle u = \tan ({e^{ - x}})$ and $\displaystyle dv = {e^{ - x}}$, but those only seem to complicate the problem.

Any pointers?

I've tried u substitution using $\displaystyle {u = {e^{ - x}}}$ ..., but those only seem to complicate the problem.
I would beg to differ on that. Your substitution should make things clearer.

$\displaystyle u = e^{-x} \implies du = -e^{-x}~dx$

$\displaystyle \int {{e^{ - x}}\tan {e^{ - x}}} dx = -\int \tan{u}~du = -\int \frac{\sin{u}}{\cos{u}}~du$

which should be straightforward.

Alternatively, you can multiply by $\displaystyle \frac{\sec{u}}{\sec{u}}.$

I don't know how to begin with this problem:

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

I've tried u substitution using $\displaystyle {u = {e^{ - x}}}$ as well as integrations by parts using $\displaystyle u = \tan ({e^{ - x}})$ and $\displaystyle dv = {e^{ - x}}$, but those only seem to complicate the problem.

Any pointers?
Hi

The substitution $\displaystyle u=e^{-x}$ works well
$\displaystyle du=-e^{-x}dx$

$\displaystyle \int {{e^{ - x}}\tan {e^{ - x}}} dx = -\int \tan u\: du$

EDIT : too late !

4. Heh... I miss the obvious too often. Thank you.