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## Homework Statement

Hi everyone. I'm trying to integrate the function

[tex]\int_0^{\pi} \frac{1}{(1+\cos{\theta}) \sqrt{1+q^2 \cos^2{\theta}+2q\cos{\theta}-p^2 \sin^2{\theta}}} d\theta[/tex]

analytically, where p and q are constants satisfying [tex] p^2 + q^2 =1 [/tex].

## Homework Equations

[tex] p^2 + q^2 =1 [/tex]

[tex] \cos{\theta} = \frac{1}{2}(e^{i\theta}+e^{-i\theta}), \sin{\theta} = \frac{1}{2i}(e^{i\theta}-e^{-i\theta}) [/tex]

Residue theorem, Cauchy's theorem, etc

## The Attempt at a Solution

It seemed to me that the only way to do this was as a contour integral, so I made the substitution

[tex] z = e^{i\theta} \rightarrow dz = iz d\theta \rightarrow d\theta = -i/z dz [/tex].

Using the above trig substitutions, I simplified and got the integral

[tex] \int \frac{1}{(1+z)^2 \sqrt{4+2(q^2-p^2)+\frac{1}{z^2}+z^2+\frac{4q}{z}+4qz} }dz[/tex].

It's still rather ugly, but I can see that there is a pole of order 2 at z=-1. I think I can neglect the roots of the stuff in the square root because each root has a factor of (-1+p^2+q^2) = 0 in the denominator, making all roots infinite. The issue now (assuming my pole analysis is ok) is to choose a contour, but because the sqrt in the denominator is not single valued I need to make a branch cut. Also, I'm integrating from 0 to pi, so that's a half circle in the complex plane. I can't see how to make a contour that includes both the branch cut and goes around the pole at z=-1. Can anyone help, or at least tell me if what I've done so far is valid? :)

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