Evaluating $\int_0^{2\pi}\frac{dt}{(a+b\cos(t))^2}=\frac{2\pi a}{(a^2-b^2)^{3/2}}$ using Residue theorem

Asked Nov 13 '19 at 16:17

Active Nov 13 '19 at 16:29

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I was solving some problems in residues, I got this problem. I tried the usual way, changing the integrating path from line segment to the unit circle by substuting $z=e^{it}$, Put in denominator i got $4$ degree polynomial, It is very difficult to proceed further. Is there any easy way to do it?

edited Nov 13 '19 at 16:29

asked Nov 13 '19 at 16:17

Madhan Kumar

I want a proof using residue theorem – Madhan Kumar Nov 13 '19 at 16:29
Then see this: https://math.stackexchange.com/q/2331632 – Nov 13 '19 at 16:32
ohh tq so much, got it – Madhan Kumar Nov 13 '19 at 16:35

Evaluating $\int_0^{2\pi}\frac{dt}{(a+b\cos(t))^2}=\frac{2\pi a}{(a^2-b^2)^{3/2}}$ using Residue theorem

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