<s>[INST] Anyone know if this integral has an analytic solution? I've come across the following integral:
$$\int_{-\pi}^{\pi}\left[\frac{1}{A-R \cos(2\theta-\phi)}\right]^{\frac{N-1}{2}}d\theta$$
I know how to approximate this integral using the Laplace method, just wondering if:
a) Does this integral have an exact answer?
b) Is there a better approximation than Laplace method for this integral? If so, under what conditions will it be better?
My thinking is that it will be a hypergeometric function (mainly because every hard integral I've come across turns out to be one of these).  Conditions (if needed) are $A>R>0$, and $N$ is an integer.
 [/INST] From Maple... $K$ and $E$ are elliptic integrals.
$$
\int_{0}^{\pi} \sqrt{\frac{1}{2 - \operatorname{cos} (2 t)}} d t = \frac{2 \sqrt{3} K \Bigl(\frac{\sqrt{3} \sqrt{2}}{3}\Bigr)}{3}
$$ $$
\int_{0}^{\pi} \frac{1}{2 - \operatorname{cos} (2 t)} d t = \frac{\pi \sqrt{3}}{3}
$$ $$
\int_{0}^{\pi} \frac{1}{2 - \operatorname{cos} (2 t)}^{\frac{3}{2}} d t = \frac{2 \sqrt{3} E \Bigl(\frac{\sqrt{3} \sqrt{2}}{3}\Bigr)}{3}
$$ $$
\int_{0}^{\pi} \bigl(2 - \operatorname{cos} (2 t)\bigr)^{(-2)} d t = \frac{2 \pi \sqrt{3}}{9}
$$ $$
\int_{0}^{\pi} \frac{1}{2 - \operatorname{cos} (2 t)}^{\frac{5}{2}} d t =  \frac{-2\sqrt{3} K \Bigl(\frac{\sqrt{3} \sqrt{2}}{3}\Bigr)}{27} + \frac{16 \sqrt{3} E \Bigl(\frac{\sqrt{3} \sqrt{2}}{3}\Bigr)}{27}
$$ $$\int_{0}^{\pi} \bigl(2 - \operatorname{cos} (2 t)\bigr)^{(-3)} d t = \frac{\pi \sqrt{3}}{6}
$$ $$
\int_{0}^{\pi} \frac{1}{2 - \operatorname{cos} (2 t)}^{\frac{7}{2}} d t = 
\frac{-32\sqrt{3} K \Bigl(\frac{\sqrt{3} \sqrt{2}}{3}\Bigr)}{405} + \frac{202 \sqrt{3} E \Bigl(\frac{\sqrt{3} \sqrt{2}}{3}\Bigr)}{405}
$$ $$
\int_{0}^{\pi} \bigl(2 - \operatorname{cos} (2 t)\bigr)^{(-4)} d t = \frac{11 \pi \sqrt{3}}{81}
$$
This suggests a reduction formula, together with the first two.
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