# Need help with a horrible integral Dont get scared

1. Nov 19, 2008

### Bkkkk

Need help with a horrible integral!! Dont get scared

1. The problem statement, all variables and given/known data
Prove that

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*$$\int$$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. Nov 19, 2008

### Dick

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.