The value of ∫limits_0^π / 2 log tan x d x, is

Let $I=\int\limits_0^{\pi / 2} \log \tan x d x$             ....(i)

Then,

$I =\int\limits_0^{\pi / 2} \log \tan \left(\frac{\pi}{2}-x\right) d x$         [Using $\int\limits_0^a f(x)dx = \int\limits_0^a f(a-x)dx$]

$\Rightarrow I =\int\limits_0^{\pi / 2} \log \cot x d x$           ....(ii)

Adding (i) and (ii), we get

$2 I=\int\limits_0^{\pi / 2}(\log \tan x+\log \cot x) d x$

$\Rightarrow 2 I=\int\limits_0^{\pi / 2} \log 1 d x=\int\limits_0^{\pi / 2} 0 d x=0 \Rightarrow I=0$