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davavsh
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∫(dθ)/(a+bcosθ)^2
I'm trying to find the above integral (from 0-2pi) using Cauchy's Residue theorem. After closing the contour and re-writing the integrant, I know that I have singularity at (-a/b)+(√(a/b)^2-1)- (double pole or is it??).
I have tried both single pole and double pole residues using the limit approach but the calculation gets very cumbersome and i don't arrive at the write answer [its (2a*pi)/(a^2-b^2)^(3/2)]. According the Mathematica Res[(-a/b)+(√(a/b)^2-1)]=b*(b^2-a^2)*((a/b)^2-1)^(1/2).
Any tips/suggestions would be appreciate on how to arrive at the result given by Mathematica (which agrees with answer given in the book).
Thanks,
Homework Equations
I'm trying to find the above integral (from 0-2pi) using Cauchy's Residue theorem. After closing the contour and re-writing the integrant, I know that I have singularity at (-a/b)+(√(a/b)^2-1)- (double pole or is it??).
The Attempt at a Solution
I have tried both single pole and double pole residues using the limit approach but the calculation gets very cumbersome and i don't arrive at the write answer [its (2a*pi)/(a^2-b^2)^(3/2)]. According the Mathematica Res[(-a/b)+(√(a/b)^2-1)]=b*(b^2-a^2)*((a/b)^2-1)^(1/2).
Any tips/suggestions would be appreciate on how to arrive at the result given by Mathematica (which agrees with answer given in the book).
Thanks,