Question

${\int }_0^{\pi /4}(\tan x+\cot x{)}^{-2}dx$

Answer 1

${\int }_0^{\frac{\pi }{4}}{(\tan x+\cot x)}^{-2}dx$

$={\int }_0^{\frac{\pi }{4}}{\left[\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right]}^{-2}$

$={\int }_0^{\frac{\pi }{4}}{\left[\frac{{\sin }^{2}x+\cos x}{\cos x\sin x}\right]}^{-2}$

$={\int }_0^{\frac{\pi }{4}}{\left(\cos x\sin x\right)}^2$

$=\frac{1}{4}[{\int }_0^{\frac{\pi }{4}}{\sin }^{2}2x$

$=\frac{1}{4}\left[{\int }_0^{\frac{\pi }{4}}\frac{1-\cos 4x}{2}\right]$

$=\frac{1}{8}\left[x-\frac{\sin 4x}{4}\right]$

$=\frac{1}{8}\left[\frac{\pi }{4}\right]$

$=\frac{\pi }{32}$