Question

Solve :-

$I={\int }_0^{\infty }\frac{dx}{e^x+e^{-x}}$

Answer 1

${\int }_0^{\infty }\frac{1}{e^x+e^{-x}}dx$
$={\int }_0^{\infty }\frac{e^x}{e^{2x}+1}dx$
Let $e^x=t$, then $e^{x}dx=dt$
When $x=0,t=1$, $x\to \infty ,t\to \infty$
$={\int }_1^{\infty }\frac{1}{t^2+1}dt$
$={\left[{\tan }^{-1}x\right]}_1^{\infty }$
$=\frac{\pi }{2}-\frac{\pi }{4}$
$=\frac{\pi }{4}$