Integral arising from an expected value problem Find $$\int_0^{\pi}\int_y^{\pi} (\pi-y)\sqrt{3+2(\cos x + \cos y+\cos (y+x))}\mathrm{d}x \mathrm{d}y.$$
I have had absolutely no trouble reducing an expected value problem to the solution of this integral. Unfortunately, the integral itself seems to be very difficult. How should I approach it?
Interestingly enough, we derive the inequality $\cos x + \cos y + \cos(x+y) \ge -3/2$ (equality holds when $x = y = \frac{2}{3}\pi$). This seems to suggest the following strategy for deriving new relations, which I will call "sideways mathematics": convert a problem to the evaluation of an algebraic expression, and interpret some propery inherent of the expression (due to where it came from) to obtain a new fact. Has anyone considered this tactic before? The only issue with this approach seems to be the lack of control over what you end up deriving.
 A: Just a hint that might be helpful.
Separate geometrically the region of integration (big triangle) into two similar triangles and a square:
$$I_1=\int_0^{\pi/2}\int_y^{\pi/2} (\pi-y)\sqrt{3+2(\cos x + \cos y+\cos (y+x))}\mathrm{d}x \mathrm{d}y$$
$$I_2=\int_{\pi/2}^{\pi}\int_y^{\pi} (\pi-y)\sqrt{3+2(\cos x + \cos y+\cos (y+x))}\mathrm{d}x \mathrm{d}y$$
$$I_3=\int_{0}^{\pi/2}\int_{\pi/2}^{\pi} (\pi-y)\sqrt{3+2(\cos x + \cos y+\cos (y+x))}\mathrm{d}x \mathrm{d}y$$
$$I=I_1+I_2+I_3$$
Substitution: $x \to \pi/2+x$ and $ y \to \pi/2+y$ in the second integral and $x \to \pi/2+x$ in the third integral gives us:
$$I_1=\int_0^{\pi/2}\int_y^{\pi/2} (\pi-y)\sqrt{3+2(\cos x + \cos y+\cos (y+x))}\mathrm{d}x \mathrm{d}y$$
$$I_2=\int_0^{\pi/2}\int_y^{\pi/2} \left(\frac{\pi}{2}-y \right)\sqrt{3-2(\sin x + \sin y+\cos (y+x))}\mathrm{d}x \mathrm{d}y$$
$$I_3=\int_0^{\pi/2}\int_0^{\pi/2} \left(\pi-y \right) \sqrt{3-2(\sin x - \cos y+\sin (y+x))}\mathrm{d}x \mathrm{d}y$$

Now all the trigonometric functions are monotone which makes it easier to apply substitutions.
I'm still not sure the closed form exists, because of the linear term in front of the root. Though it should be possible to integrate w.r.t. $x$.
