Can Somebody please solve this integral for me (showing their steps)??

**skinsdomination09** · November 14th 2012, 04:18 PM

e ^ -x * tan(e^ -x)

Sorry I would try and explain everything I've tried but i'm on pre-test overload going through tons of problems.

I thought maybe u would be e to the -x but I couldn't get it to work.

thanks

**MacstersUndead** · November 14th 2012, 04:21 PM

Let
u = tan(e^-x) and v' = e^-x and try to use integration by parts. [I suggest v' as I do because it's easier to integrate. You might have to use a different method]

$\int u(x) v'(x) \, dx = u(x) v(x) - \int u'(x) v(x) \, dx$

EDIT: You can also use u substitution and let u = e^-x

**Soroban** · November 14th 2012, 05:00 PM

Hello, skinsdomination09!

Your game plan was a good one . . . It should have worked.

$\int e^{\text{-}x}\tan(e^{\text{-}x})\,dx$

We have: . $\int \tan(e^{\text{-}x})\,(e^{\text{-}x}dx)$

Let $u \,=\,e^{\text{-}x} \quad\Rightarrow\quad du \,=\,\text{-}e^{\text{-}x}dx \quad\Rightarrow\quad e^{\text{-}x}dx \,=\,\text{-}du$

Substitute: . $\int \tan u (\text{-}du) \;=\;-\int\tan u\,du \;=\;\ln|\cos u| + C$

Back-substitute: . $\ln|\cos(e^{\text{-}x})| + C$

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