Question

If $I=\underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{dx}{(1+e^{\sin x})(2-\cos 2x)},$ then the value of $\left[I\right]$ is
(where $[.]$ represents the greatest integer function)

Answer 1

$I=\underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{dx}{(1+e^{\sin x})(2-\cos 2x)}...\left(1\right)$

By using $\underset{a}{\overset{b}{\int }}f\left(x\right)dx=\underset{a}{\overset{b}{\int }}f(a+b-x)dx$

$I=\underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{e^{\sin x}dx}{(1+e^{\sin x})(2-\cos 2x)}...\left(2\right)$

Adding eqn $\left(1\right)$ and $\left(2\right),$ we get
$2I=\underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{dx}{2-\cos 2x}$

$2I=\underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{dx}{1+2{\sin }^{2}x}$

Since, ${\sin }^{2}x$ is an even function, therefore we have
$I=\underset{0}{\overset{\pi /2}{\int }}\frac{dx}{1+2{\sin }^{2}x}$

$I=\underset{0}{\overset{\pi /2}{\int }}\frac{{\sec }^{2}xdx}{2{\tan }^{2}x+{\sec }^{2}x}$

$I=\underset{0}{\overset{\pi /2}{\int }}\frac{{\sec }^{2}xdx}{3{\tan }^{2}x+1}$

Put $\tan x=t\Rightarrow {\sec }^{2}xdx=dt$
$I=\underset{0}{\overset{\infty }{\int }}\frac{dt}{3t^2+1}$
$=\frac{1}{\sqrt{3}}[{\tan }^{-1}\left(\sqrt{3}t\right){]}_0^{\infty }$
$=\frac{\pi }{2\sqrt{3}}$
$=\frac{3.14}{2\times 1.73}<1$

$\therefore \left[I\right]=0$