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.
 A: As joriki points out, you can drop the $\phi$. Further assuming that $A > R$ so that the denominator never vanishes, the integral can be evaluated by resorting to complex analysis. Let $N = 2m + 1$
$$I = \int_{-\pi}^{\pi} \left(\frac1{A - R \cos(2 \theta)} \right)^{m} d \theta$$
Let $z = e^{i \theta}$, then $dz = i z d \theta$. Hence, the integral becomes
$$I = \oint_{|z| = 1} \frac{dz}{iz \left(A - \frac{R}{2} \left(z^2 + \frac1{z^2} \right) \right)^m}$$
$$I = \oint_{|z| = 1} \frac{2^m z^{2m} dz}{iz \left(2Az^2 - R \left(z^4 + 1 \right) \right)^m}$$
$$I = -i \left(- \frac{2}{R} \right)^m \oint_{|z| = 1} \frac{z^{2m-1} dz}{\left(\left(z^4 + 1 \right) - 2 \frac{A}{R} z^2 \right)^m}$$
And evaluate the integral using residue theorem.
The poles of the integrand are at $$z= \pm \sqrt{\frac{A \pm \sqrt{A^2 - R^2}}{R}}$$
Since $A>R$, the poles within the contour lie at $$z = \pm \sqrt{\frac{A - \sqrt{A^2 - R^2}}{R}}$$
A: 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.
A: If $N$ is odd, then the exponent is even, so you've got a rational function of sine and cosine, so the Weierstrass substitution should work: http://en.wikipedia.org/wiki/Weierstrass_substitution
