Given,

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

On applying property

$$\int_a^b {f(x)dx} = \int_a^b {f(a + b - x)dx} $$, we get

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

On adding integrals (i) and (ii), we get

$$2I = {2 \over \pi }\int_{ - \pi /4}^{\pi /4} {{{dx} \over {2 - \cos 2x}}} $$

$$ \Rightarrow $$ $$I = {1 \over \pi }\int_{ - \pi /4}^{\pi /4} {{{dx} \over {2 - {{1 - {{\tan }^2}x} \over {1 + {{\tan }^2}x}}}}} $$

$$\left[ {as\,\cos \,2x = {{1 - {{\tan }^2}x} \over {1 + {{\tan }^2}x}}} \right]$$

= $${2 \over \pi }\int_0^{\pi /4} {{{{{\sec }^2}x} \over {1 + 3{{\tan }^2}x}}dx} $$

[$$ \because $$ $${{{{{\sec }^2}x} \over {1 + 3{{\tan }^2}x}}}$$ is even function]

Put $$\sqrt 3 \tan \,x = t \Rightarrow \sqrt 3 {\sec ^2}dx = dt$$, and at x = 0, t = 0 and at x = $$\sqrt 3 $$, t = $$\sqrt 3 $$

So, $$I = {2 \over \pi }\int_0^{\sqrt 3 } {{1 \over {\sqrt 3 }}{{dt} \over {1 + {t^2}}}} = {2 \over {\sqrt 3 \pi }}[{\tan ^{ - 1}}t]_0^{\sqrt 3 }$$

$$ = {2 \over {\sqrt 3 \pi }}\left( {{\pi \over 3}} \right) = {2 \over {3\sqrt 3 }} \Rightarrow 27{I^2} = 4.00$$