Evaluate $$\int_{0}^{2\pi}\int_{0}^{\pi} {\cos\phi \sin\phi \over \sqrt{R^2+r^2-2Rr(\cos\phi \cos\theta+\sin\phi \sin\theta \cos\psi )}} d\phi\ d\psi$$ where $R,r,\theta$ are all constants.
Sorry for all those distracting constants. (This integral came up from physics calculation.) My first idea was substitution $$\cos\phi \cos\theta+\sin\phi \sin\theta \cos\psi = 1+{\sin^2{\eta}\over 2rR}$$ but I don't think this approach is fruitful.
Even a simple hint about variable substitution will help me a lot. Thank you.
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With AlexR's help, I did one integration with respect to $\psi$, so this is the new integral with respect to $\phi$ $$\int_{0}^{2\pi}{\cos \phi \sin \phi K({2Rr\sin \theta \sin\phi \over R^2+r^2 -2Rr\cos(\phi-\theta)})\over\sqrt{R^2+r^2 -2Rr\cos(\phi-\theta)}}d\phi$$ where $K$ is the complete elliptic integral of the first kind.
Now I'm terrified with the presence of special function in the integrand. Can this integral even be done?
By spherical symmetry, the integral should be independent of $\theta$, so set$\theta=0$ which gives $\pi$ times an integral which vanishes by symmetry. Hence the value is 0.