The value of the integral $\int\limits_0^{\pi / 2} \log |\tan x+\cot x| d x$ is

$\pi \log 2$

Let

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

$\Rightarrow I =\int\limits_0^{\pi / 2} \log \left|\frac{\sin ^2 x+\cos ^2 x}{\sin x \cos x}\right| d x$

$\Rightarrow I =\int\limits_0^{\pi / 2} \log \left(\frac{1}{\sin x \cos x}\right) d x$         $\left[\begin{array}{l} ∵ \sin x \text { and } \cos x \text { are } \\ \text { positive in first quadrant } \end{array}\right]$

$\Rightarrow I=-\int\limits_0^{\pi / 2} \log \sin x d x-\int\limits_0^{\pi / 2} \log \cos x d x$

$\Rightarrow I=-(-\pi / 2 \log 2)-(-\pi / 2 \log 2)=\pi \log 2$