If $I=\frac{2}{\pi}\int_{-\pi/4}^{\pi/4}\frac{dx}{(1+e^{\sin x})(2-\cos 2x)}$ , then $27\,I^{2}$ equals ___,

$I = \frac{2}{\pi}\int_{-\pi/4}^{\pi/4}\frac{dx}{(1+e^{\sin x})(2-\cos 2x)}$

$\text{Find } 27\,I^{2}$

$1.\ \text{symmetry kills } e^{\sin x}$

$2.\ \text{substitute } t=\tan x$

$3.\ \text{find } I \text{ and } 27\,I^{2}$

$\int_{-a}^{a} f(x)\,dx = \int_{-a}^{a} f(-x)\,dx$

$J = \int_{-\pi/4}^{\pi/4}\frac{dx}{(1+e^{\sin x})(2-\cos 2x)}$

$2J = \int_{-\pi/4}^{\pi/4}\frac{1}{2-\cos 2x}\left[\frac{1}{1+e^{\sin x}}+\frac{1}{1+e^{-\sin x}}\right]dx$

$\frac{1}{1+e^{s}}+\frac{1}{1+e^{-s}} = \frac{1+e^{s}}{1+e^{s}} = 1$

$2J = \int_{-\pi/4}^{\pi/4}\frac{dx}{2-\cos 2x} \ \Rightarrow\ J = \int_{0}^{\pi/4}\frac{dx}{2-\cos 2x}$

$t=\tan x,\quad dx=\frac{dt}{1+t^{2}},\quad \cos 2x=\frac{1-t^{2}}{1+t^{2}}$

$J = \int_{0}^{1}\frac{dt}{1+3t^{2}}$

$J = \frac{1}{\sqrt{3}}\Big[\tan^{-1}(\sqrt{3}t)\Big]_0^1 = \frac{1}{\sqrt{3}}\cdot\frac{\pi}{3} = \frac{\pi}{3\sqrt{3}}$

$I = \frac{2}{\pi}\cdot J = \frac{2}{\pi}\cdot\frac{\pi}{3\sqrt{3}} = \frac{2}{3\sqrt{3}}$

$I^{2} = \frac{4}{27} \Rightarrow 27\,I^{2} = 4$

$\boxed{27\,I^{2} = 4}$