Suppose the integral $$ \tag 1 I = \int \limits_{-\pi}^{\pi}dx \int \limits_{-\pi}^{\pi}\frac{dy}{\tau - \cos (2x) -2\cos(x)\cos(y)}, \quad t > 3 $$ How to evaluate it in terms of elliptic integral?
My attemption.
I made the substitution $$ \cos(x) = t, \quad \cos(y) = k $$ My integral then took the form $$ I = 4\int \limits_{-1}^{1}\int \limits_{-1}^{1}\frac{dxdy}{\sqrt{1-x^{2}}\sqrt{1-y^{2}}}\frac{1}{\tau +1 -2x^2-2xy} = $$ $$ =4\pi \int \limits_{-1}^{1}\frac{dx}{\sqrt{(1-x^2)((\tau+1-2x^2)^{2} - 4x^2)}} = |x = \sqrt{u}| $$ $$ =4\pi \int \limits_{0}^{1}\frac{du}{\sqrt{(1-u)u((\tau + 1 - 2u)^2 - 4u)}} $$ But I don't see how it reduces to elliptic integral. Could You help me?
P.S. The elliptic integral from the answer has terrible argument $$ k = \frac{4(2\tau+3)^{\frac{1}{4}}}{\sqrt{(\sqrt{2\tau+3}-1))^{3}(\sqrt{2\tau+3}+3})} $$
An edit. The integral is taken.