Solving an Integral Help

**summermagic** · March 29th 2009, 01:23 PM

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

I'm not sure how to start this problem so any help would be appreciated.

**stapel** · March 29th 2009, 01:49 PM

Originally Posted by summermagic

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

I didn't see anything useful, so I plugged your integral into The Integrator....

Working backwards, you can easily confirm that $arctan\left(\sqrt{e^{2x}\, -\, 1}\right)$ does indeed differentiate back to your integrand, but I'm not seeing how you're supposed to come up with this "going frontways"....

**Soroban** · March 29th 2009, 01:54 PM

Hello, summermagic!

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

It can be done with a Trig Substitution: . $e^x \,=\,\sec\theta$
. . but here is a back-door approach.

Divide top and bottom by $e^x$

. $\int\frac{\dfrac{dx}{e^x}}{\dfrac{\sqrt{e^{2x}-1}}{e^x}} \;=\;\int \frac{e^{\text{-}x}\,dx}{\sqrt{\dfrac{e^{2x}-1}{e^{2x}}}} \;=\;\int\frac{e^{\text{-}x}dx}{\sqrt{1 - e^{\text{-}2x}}}$

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

Substitute: . $\int\frac{-du}{\sqrt{1-u^2}} \;=\;\arccos u + C$

Back-substitute: . $\arccos\left(e^{\text{-}x}\right) + C$

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