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$\int_{0}^{\pi / 2} \log (\tan x) d x=$
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Let $\quad I=\int_{0}^{\pi / 2} \log (\tan x) d x$
$\Rightarrow \quad I=\int_{0}^{\pi / 2} \log \tan \left(\frac{\pi}{2}-x\right) d x$
$\Rightarrow \quad I=\int_{0}^{\pi / 2} \log \cot x d x$
$\Rightarrow \quad I=\int_{0}^{\pi / 2} \log \frac{1}{\tan x} d x$
$\Rightarrow \quad I=-\int_{0}^{\pi / 2} \log \tan x d x$
$\Rightarrow \quad I=-I \Rightarrow 2 I=0 \Rightarrow I=0$
$\Rightarrow \quad I=\int_{0}^{\pi / 2} \log \tan \left(\frac{\pi}{2}-x\right) d x$
$\Rightarrow \quad I=\int_{0}^{\pi / 2} \log \cot x d x$
$\Rightarrow \quad I=\int_{0}^{\pi / 2} \log \frac{1}{\tan x} d x$
$\Rightarrow \quad I=-\int_{0}^{\pi / 2} \log \tan x d x$
$\Rightarrow \quad I=-I \Rightarrow 2 I=0 \Rightarrow I=0$
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