Question

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

Answer 1

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

Take $x$ as $\frac{\pi }{2}-x$ we get

$I={\int }_0^{\frac{\pi }{2}}\log \left(\tan (\frac{\pi }{2}-x)\right)dx$

$I={\int }_0^{\frac{\pi }{2}}\log \left(\cot x\right)dx$ ........$\left(2\right)$

Adding $\left(1\right)$ and $\left(2\right)$ we get

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

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

$2I={\int }_0^{\frac{\pi }{2}}\left[\log \left(\tan x\cot x\right)\right]dx$

$2I={\int }_0^{\frac{\pi }{2}}\left[\log 1\right]dx$

$2I={\int }_0^{\frac{\pi }{2}}0dx$

$I=\frac{1}{2}\times 0=0$

$\therefore I=0$