How to integrate $\int_0^{2\pi} \frac{1}{(k\cos\theta + 1)^2} d\theta$

Asked Aug 31 '19 at 06:00

Active Aug 31 '19 at 06:00

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I have the following identity in a derivation I'm trying to understand:

$\int_0^{2\pi} \frac{1}{(k\cos\theta + 1)^2} d\theta = \frac{2\pi}{(1-k^2)^\frac{3}{2}}$

and I cannot see how this is true. I'm sure it is - I just can't figure out how to integrate the LHS. I've tried using tangent half-angle substitution, and although that gave me something I could integrate, the definite integral ended up being $0 - 0 = 0$, which wasn't entirely helpful...

Can anyone help?

asked Aug 31 '19 at 06:00

JeffR

1

It should be $-1<k<1.$ Otherwise the integral diverges. – Michael Rozenberg Aug 31 '19 at 06:03
Have you checked this posting? How to evaluate $\int_0^{2\pi}\frac{1}{(1+a\cos {\theta})^2},d\theta$ without contour integration? – Sangchul Lee Aug 31 '19 at 06:10
@SangchulLee No, I hadn't seen that - and it wasn't presented as a similar question. But it does answer my question - thanks. – JeffR Aug 31 '19 at 06:21

How to integrate $\int_0^{2\pi} \frac{1}{(k\cos\theta + 1)^2} d\theta$

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