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Math Help - Solving an Integral Help

  1. #1
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    Solving an Integral Help

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

    I'm not sure how to start this problem so any help would be appreciated.
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  2. #2
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    Question

    Quote Originally Posted by summermagic View Post
    \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"....
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  3. #3
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    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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