# How do you integrate int e^-xtan(e^-x)dx?

Dec 13, 2016

$\int {e}^{-} x \tan \left({e}^{-} x\right) \mathrm{dx} = \ln \left(\left\mid \cos \right\mid \left({e}^{-} x\right)\right) + C$

#### Explanation:

$\int {e}^{-} x \tan \left({e}^{-} x\right) \mathrm{dx}$

First, let $u = {e}^{-} x$. Differentiating this shows that $\mathrm{du} = - {e}^{-} x \mathrm{dx}$.

$= - \int \tan \left({e}^{-} x\right) \left(- {e}^{-} x\right) \mathrm{dx}$

$= - \int \tan \left(u\right) \mathrm{du}$

This is a standard integral, but we can show how to integrate it by rewriting tangent as sine divided by cosine:

$= - \int \sin \frac{u}{\cos} \left(u\right) \mathrm{du}$

Now, let $v = \cos \left(u\right)$. This implies that $\mathrm{dv} = - \sin \left(u\right) \mathrm{du}$.

$= \int \frac{- \sin \left(u\right)}{\cos} \left(u\right) \mathrm{du}$

$= \int \frac{\mathrm{dv}}{v}$

This is also a standard (and very important) integral:

$= \ln \left(\left\mid v \right\mid\right) + C$

$= \ln \left(\left\mid \cos \right\mid \left(u\right)\right) + C$

$= \ln \left(\left\mid \cos \right\mid \left({e}^{-} x\right)\right) + C$