Exact expression of a trigonometric integral Let $a>2$ be a real number and consider the following integral
$$
I(a)=\int_0^\pi\int_0^\pi \frac{\sin^2(x)\sin^2(y)}{a+\cos(x)+\cos(y)} \mathrm{d}x\,\mathrm{d}y
$$

My question. Does there exist a closed-form expression of $I(a)$?

Some comments. Since $a-2<a+\cos(x)+\cos(y)<a+2$ and $\int_0^\pi \int_0^\pi \sin^2(x)\sin^2(y)\ \mathrm{d}x\, \mathrm{d}y=\frac{\pi^2}{4}$, we have the following bounds
$$
\frac{\pi^2}{4(a+2)} < I(a) < \frac{\pi^2}{4(a-2)},
$$ 
however I didn't manage to find an exact expression for $I(a)$. Any help is welcome!

We use the identities
$$
2\sin \phi \sin \psi=\cos(\phi-\psi)-\cos(\phi+\psi)
$$
and
$$
\cos \phi+\cos \psi=2\cos\left(\frac{\phi+\psi}{2}\right)\cos\left(\frac{\phi-\psi}{2}\right)
$$
to get
$$
I=\int^{\pi}_{0}\int^{\pi}_{0}\frac{\sin^2 x \sin^2 y}{a+\cos x+\cos y}dxdy=
$$
$$
\frac{1}{4}\int^{\pi}_{0}\int^{\pi}_{0}\frac{(\cos(x-y)-\cos(x+y))^2}{a+2\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)}dxdy
$$
Now we use $\cos \phi=2\cos^2(\phi/2)-1$ and arrive to
$$
I=\int^{\pi}_{0}\int^{\pi}_{0}\frac{(\cos^2\left(\frac{x-y}{2}\right)-\cos^2\left(\frac{x+y}{2}\right))^2}{a+2\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)}dxdy=
$$
$$
=a^{-1}\int^{\pi}_{0}\int^{\pi}_{0}\frac{(\cos^2 A-\cos^2 B)^2}{1+2/a \cos A\cos B}dAdB.
$$
Then set $t=\cos A$, $s=\cos B$ and use
$$
\frac{d\left(\cos^{(-1)}(x)\right)}{dx}=-\frac{1}{\sqrt{1-x^2}}
$$
to get
$$
I=1/a\int^{1}_{-1}\int^{1}_{-1}\frac{(t^2-s^2)^2}{1+(2/a)ts}\frac{1}{\sqrt{1-t^2}}\frac{1}{\sqrt{1-s^2}}dtds
$$
If $|2/a|\leq  1$ and $|t|,|s|<1$, we expand $(1+(2/a)ts)^{-1}$ into $\sum^{\infty}_{k=0}(-1)^k(2/a)^kt^ks^k$ we can split the two integrals (that of $t$ and that of $s$). We set
$$
I_k:=1/a\int^{1}_{-1}\int^{1}_{-1}(t^2-s^2)^2(-1)^k(2/a)^kt^ks^k\frac{1}{\sqrt{1-t^2}}\frac{1}{\sqrt{1-s^2}}dtds.
$$
Then $I_{2k+1}=0$  and
$$
I_{2k}=\frac{2^{2k}(1/a)^{2k+1}\pi\Gamma\left(\frac{2k+1}{2}\right)\Gamma\left(\frac{2k+3}{2}\right)}{\Gamma\left(k+2\right)\Gamma(k+3)}.
$$
Hence summing
$$
I=I(a)=\sum^{\infty}_{k=0}I_{2k}=\frac{a\pi^2}{2}-\frac{a\pi^2}{2}{}_2F_1\left(\frac{-1}{2},\frac{1}{2};2;\frac{4}{a^2}\right)=
$$
$$
=\frac{a\pi^2}{2}-\frac{a(a^2+4)\pi}{6}E\left(4a^{-2}\right)+\frac{a(a^2-4)\pi}{6}K(4a^{-2}),
$$
where $K(x)$,$E(x)$ are the complete elliptic integrals of the first and second kind. Examples of evaluations can be given if we consider the singular modulus $k_r$. For example with $a=2\sqrt{2}$, we get $k_1=1/2$ and we have
$$
I\left(2\sqrt{2}\right)=\sqrt{2}\pi^2-2\sqrt{2\pi}\Gamma\left(\frac{3}{4}\right)^2-\frac{8}{3}\sqrt{2\pi}\Gamma\left(\frac{5}{4}\right)^2
$$
...etc
see elliptic integrals singular valus
