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
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
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
Hello, skinsdomination09!
Your game plan was a good one . . . It should have worked.
$\displaystyle \int e^{\text{-}x}\tan(e^{\text{-}x})\,dx$
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: .$\displaystyle \ln|\cos(e^{\text{-}x})| + C$