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Thread: Just a hyperbolic integral

  1. #1

    Red face Just a hyperbolic integral

    I can't figure this one out,

    integral of 1/(sqrt(5-e^(2x))


    I don't see where I can make a substitution seeing as there is no e^x in the numerator,

    Thanks in advance guys.

  2. #2
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    Quote Originally Posted by Mohamed Abdelkhaliq View Post
    I can't figure this one out,

    integral of 1/(sqrt(5-e^(2x))


    I don't see where I can make a substitution seeing as there is no e^x in the numerator,

    Thanks in advance guys.
    No guarantees but just maybe
    u = [tex]\frac{1}{\sqrt{5 - e^{2x}}}[/tex]
    so that
    e2x = 5 - u-2
    Then use partial fractions

  3. #3
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    Or: start with [tex]u= e^{x}[/tex] so that [tex]du= e^x dx[/tex] or [tex]dx= \frac{1}{u}du[/tex], [tex]\sqrt{5- e^{2x}}= \sqrt{5- u^2}[/tex] and
    [tex]\int \frac{dx}{\sqrt{5- e^{2x}}}= \int \frac{du}{u\sqrt{5- u^2}}[/tex].

    Now use a trig substitution: [tex]u= \sqrt{5} sin(t)[/tex] so that [tex]du= \sqrt{5}cos(t)dt[/tex], [tex]\sqrt{5- u^2}= \sqrt{5(1- sin^2(t))}= \sqrt{5}cos(t)[/tex] and
    [tex]\int \frac{du}{u\sqrt{5- u^2}}[/tex][tex]= \int \frac{\sqrt{5}cos(t)dt}{(\sqrt{5} sin(t))(\sqrt{5}cos(t))}=[/tex][tex] \frac{\sqrt{5}}{5}\int cosec(t)dt[/tex]

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