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

• Nov 14th 2012, 04:18 PM
skinsdomination09
Can Somebody please solve this integral for me (showing their steps)??
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 :)
• Nov 14th 2012, 04:21 PM
Re: Can Somebody please solve this integral for me (showing their steps)??
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]

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

EDIT: You can also use u substitution and let u = e^-x
• Nov 14th 2012, 05:00 PM
Soroban
Re: Can Somebody please solve this integral for me (showing their steps)??
Hello, skinsdomination09!

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

Quote:

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

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

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

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

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