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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.

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belial: to enclose the square root of an expression, use braces: sqrt{expression}. Let me know if it got the parentheses wrong in your second expression! – amWhy Dec 4 '12 at 14:44
oh, that looks much better now! thank you! – Belial Dec 4 '12 at 14:46

1 Answer 1

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/

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well, if you substitute with $u=e^{-x}$ you also get $\frac{du}{dx}=-e^{-x} \Rightarrow -du = e^{-x} dx$ so you get $$ -2 \pi \int \sqrt{1+ u^2} du $$ if we set $u = tan(t)$ then we'll get $$ -2 \pi \int \frac{\sqrt{1+ tan(t)^2}}{cos(t)^2} dt $$ But how is this easier than the previous one? – Belial Dec 4 '12 at 15:21
@Belial: $1+\tan^2 x = \sec^2 x$. – Javier Dec 4 '12 at 15:55
@Belial: fixed. – N. S. Dec 4 '12 at 19:43

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