IF $I=\frac{2}{\pi }\underset{-\pi /4}{\overset{\pi /4}{\int }}\frac{dx}{\left(1+e^{\sin x}\right)\left(2-\cos 2x\right)}$ then the value of $27I^2$ is 

$\begin{array}{l}I=\frac{2}{\pi }\underset{-\pi /4}{\overset{\pi /4}{\int }}\frac{dx}{\left(1+e^{\sin x}\right)\left(2-\cos 2x\right)}\\ I=\frac{2}{\pi }\underset{-\pi /4}{\overset{\pi /4}{\int }}\frac{e^{\sin x}}{\left(1+e^{\sin x}\right)\left(2-\cos 2x\right)}dx \left(\sin ce {\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left(a+b-x\right)dx\right)\\ \text{Adding the above both integrals}\\ \text{2I=}\frac{2}{\pi }\underset{-\pi /4}{\overset{\pi /4}{\int }}\frac{dx}{\left(2-\cos 2x\right)}\\ I=\frac{1}{\pi }\underset{-\pi /4}{\overset{\pi /4}{\int }}\frac{{\sec }^{2}xdx}{1+3{\tan }^{2}x}\\ =\frac{2}{3\sqrt{3}}\\ 27I^2=4\end{array}$