Question

$\underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{e^{\left|\sin x\right|}\cos x}{(1+e^{\tan x})}dx$ is equal to

Answer 1

Let $I=\underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{e^{\left|\sin x\right|}\cos x}{(1+e^{\tan x})}dx$
We know that,
$\underset{-a}{\overset{a}{\int }}f\left(x\right)dx=\underset{0}{\overset{a}{\int }}\left(f\right(x)+f(-x\left)\right)dx$
$=\underset{0}{\overset{\pi /2}{\int }}(\frac{e^{\left|\sin x\right|}\cos x}{1+e^{\tan x}}+\frac{e^{\left|\sin x\right|}\cos x}{1+e^{-\tan x}})dx$
$=\underset{0}{\overset{\pi /2}{\int }}{e}^{\left|\sin x\right|}\cos xdx$
$=\underset{0}{\overset{\pi /2}{\int }}{e}^{\sin x}\cos xdx$
$=[e^{\sin x}{]}_0^{\pi /2}=e-1$