0
$\begingroup$

I want to compute the integral $2\pi\int f(x) \sqrt{1+f'(x)^2} dx$ where $f(x)=\dfrac{1}{e^x}$.

I used maple and I found that the answer is: $\pi e^{-2x} \left[e^{2x} \arctan\left(\sqrt{e^{2x}-1}\right) - \sqrt{e^{2x}-1}\right] $ but I can't find a way to prove it on the paper. Any help would be apreciated.

  • 0
    oh, that looks much better now! thank you!2012-12-04

1 Answers 1

3

Note that $f(x) =e^{-x}$ thus $f'(x)=-e^{-x}$.

Your integral is

$2\pi\int e^{-x} \sqrt{1+e^{-2x}} dx$

After the substitution $u=e^{-x}$ you get

$-2 \pi \int \sqrt{1+u^2}du$

This is a standard problem for trig substitution. Set $u=\tan(t)$ and you are done/

  • 0
    @Belial: fixed.2012-12-04