QJEEAdv19Ma3d2f: JEE Advanced 2019 Mathematics - Class 12 JEEAdv19Ma3d2f Solution: If , then 27I 2 equals .................

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

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