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Need help with a horrible integral Dont get scared

  1. Nov 19, 2008 #1
    Need help with a horrible integral!! Dont get scared

    1. The problem statement, all variables and given/known data
    Prove that
    [​IMG]

    2. Relevant equations
    Cosh2-Sinh2=1


    3. The attempt at a solution

    I have used the subsitution r=2GM/c2Cosh2(x) and the integral is simplified to:

    4GM/c2*[tex]\int[/tex]Cosh2(x)dx

    Integrating was done in Mathematica but I can't do the simplification needed for the equation to look like the answer.
     
  2. jcsd
  3. Nov 19, 2008 #2

    Dick

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    Re: Need help with a horrible integral!! Dont get scared

    The integral of cosh(x)^2 is 2*cosh(x)*sinh(x)/4+x/2. The cosh(x)*sinh(x) is the first term in your answer, yes? The x/2 part is the hard one. You've got r~cosh(x)^2 (I'll ignore the constant factors). So x=arccosh(r^(1/2)). arccosh(a)=ln(a+sqrt(a^2-1)). Doesn't that look like the second term? You can throw away the third term. It's just a constant.
     
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