Question

${\int }_0^{\pi /2}\log (\tan x+\cot x)dx=\pi \left(\log 2\right)$

Answer 1

$I={\int }_0^{\pi /2}\log (\tan x+\cot x)dx$

$\Rightarrow I={\int }_0^{\pi /2}\log (\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x})dx$

$\Rightarrow I={\int }_0^{\pi /2}\log \left(\frac{{\cos }^{2}x+{\sin }^{2}x}{\sin x\cos x}\right)dx$

$I={\int }_0^{\pi /2}\log \left(\frac{1}{\sin x\cos x}\right)dx$

$I=-{\int }_0^{\pi /2}\log \left(\sin x\cos x\right)dx$

$I=-{\int }_0^{\pi /2}\left(\log \left(2\sin x\cos x\right)-\log x\right)$

$2I_1={\int }_0^{\pi /2}\log \sin x\cos xdx$

$2I_1={\int }_0^{\pi /2}\left(\log 2\sin x\cos xdx-\log 2dx\right)$

$2I_1={\int }_0^{\pi /2}\log \sin 2xdx-\frac{\pi }{2}\log 2$

$2I_1=I_1-\frac{\pi }{2}\log 2\left[from\left(3\right)\right]$

$I_1=\frac{-\pi }{2}\log 2-\left(\frac{-\pi }{2}\log 2\right)$ [ from (4) ]

$\Rightarrow I=\pi \log 2=RHS$

Hence, the answer is $\pi \log 2.$