Let $k>1$. How to evaluate: $$I=\int_{0}^{2\pi}\frac{d\theta}{\sqrt{k-\cos\theta}}$$ Since $\cos\theta$ is even around $\theta=\pi$, we have: $$I=2\int_{0}^{\pi}\frac{d\theta}{\sqrt{k-\cos\theta}}$$ I use $x=\cos\theta$, then: $dx=-\sqrt{1-x^2}d\theta$: $$I=2\int_{-1}^{1}\frac{dx}{\sqrt{1-x^2}\sqrt{k-x}}$$ and then stuck...
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4https://en.wikipedia.org/wiki/Elliptic_integral – Claude Leibovici Nov 23 '21 at 06:52
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When $k=1$, it's a quick solution using the half angle formula. For $k \neq 1$, it's not an elementary integral I think. – Deepak Nov 23 '21 at 07:01
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Byrd and Friedman 291.00 immediately gives the answer as $$\frac4{\sqrt{1+k}}K\left(m=\frac2{1+k}\right)$$ where $K$ is the complete first-kind elliptic integral and $m$ indicates the parameter (used by Mathematica/mpmath). – Parcly Taxel Nov 23 '21 at 16:18