Question

If
$I=\frac{2}{\pi }\underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\int }}\frac{dx}{(1+e^{\sin x})(2-\cos 2x)}$
then $27I^2$ equals

Answer 1

$I=\frac{2}{\pi }\underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\int }}\frac{dx}{(1+e^{\sin x})(2-\cos 2x)}\cdots \left(i\right)$
(Using $\underset{b}{\overset{a}{\int }}f\left(x\right)dx=\underset{b}{\overset{a}{\int }}f(a+b-x)dx$)
$I=\frac{2}{\pi }\underset{\frac{-\pi }{4}}{\overset{\frac{\pi }{4}}{\int }}\frac{dx}{(1+e^{-\sin x})(2-\cos 2x)}$
$I=\frac{2}{\pi }\underset{\frac{-\pi }{4}}{\overset{\frac{\pi }{4}}{\int }}\frac{e^{\sin x}dx}{(1+e^{\sin x})(2-\cos 2x)}\cdots \left(ii\right)$
Adding equation $\left(i\right)$ and $\left(ii\right)$, we get-
$2I=\frac{2}{\pi }\underset{\frac{-\pi }{4}}{\overset{\frac{\pi }{4}}{\int }}\frac{dx}{(2-\cos 2x)}$
$\because \underset{-a}{\overset{a}{\int }}f\left(x\right)dx=2\underset{0}{\overset{a}{\int }}f\left(x\right)dx\Rightarrow Evenfunction$
$2I=\frac{4}{\pi }\underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{dx}{(1+2{\sin }^{2}x)}$
$I=\frac{2}{\pi }\underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{{\sec }^{2}xdx}{(se{c}^{2}x+2{\tan }^{2}x)}$
$I=\frac{2}{\pi }\underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{{\sec }^{2}xdx}{(1+{\tan }^{2}x+2{\tan }^{2}x)}$
Put $\tan x=u\Rightarrow {\sec }^{2}xdx=du$
$I=\frac{2}{\pi }\underset{0}{\overset{1}{\int }}\frac{du}{(1+3u^2)}=[\frac{2}{\sqrt{3}\pi }{\tan }^{-1}\left(\sqrt{3}u\right){]}_0^1$
$I=\frac{2}{\sqrt{3}\pi }\times (\frac{\pi }{3}-0)=\frac{2}{3\sqrt{3}}$
$27I^2=27\times (\frac{2}{3\sqrt{3}}{)}^2=4$