${\int }_0^{\pi /2}\log (\tan x)dx$ is equal to:

Let $I={\int }_0^{\pi /2}\log (\tan x)dx$ --- (1)Using the property ${\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx$:
$$I={\int }_0^{\pi /2}\log (\tan (\frac{\pi }{2}-x))dx$$
$$I={\int }_0^{\pi /2}\log (\cot x)dx$$
--- (2)Adding (1) and (2):
$$2I={\int }_0^{\pi /2}[\log (\tan x)+\log (\cot x)]dx$$
$$2I={\int }_0^{\pi /2}\log (\tan x\cdot \cot x)dx$$
$$2I={\int }_0^{\pi /2}\log (1)dx$$
Since $\log (1)=0$:
$$2I=0⟹I=0$$