Let $\mathrm{I}=\int_{0}^{\pi / 2}(\log \tan \mathrm{x}) \cdot \sin 2 \mathrm{xdx} \ldots(\mathrm{i})$
$\mathrm{I}=\int_{0}^{\pi / 2} \log \tan \left(\frac{\pi}{2}-\mathrm{x}\right) \sin 2\left(\frac{\pi}{2}-\mathrm{x}\right) \mathrm{dx}$
$\left[\because \int_{0}^{\mathrm{a}} \mathrm{f}(\mathrm{x}) \mathrm{dx}=\int_{0}^{\mathrm{a}} \mathrm{f}(\mathrm{a}-\mathrm{x}) \mathrm{dx}\right]$
$\begin{array}{r}
\Rightarrow \quad \mathrm{I}=\int_{0}^{\pi / 2} \log \cot \mathrm{x} \cdot \sin 2 \mathrm{x} \mathrm{dx} ...(ii) \\
{[\because \sin (\pi-2 \mathrm{x})=\sin 2 \mathrm{x}]}
\end{array}$
On adding eqs (i) and (ii), we get
$\begin{array}{c}
2 \mathrm{I} \int_{0}^{\pi / 2} \log \tan \mathrm{x} \cdot \sin 2 \mathrm{x} \mathrm{dx}+\int_{0}^{\pi / 2} \log \cot \mathrm{x} \sin 2 \mathrm{x} \mathrm{dx} \\
=\int_{0}^{\pi / 2} \sin 2 x \log (\tan x \cdot \cot x) d x \\
{[\because \log \mathrm{m}+\log \mathrm{n}=\log (\mathrm{m} \cdot \mathrm{n})]}
\end{array}$
$\begin{aligned}
&=\int_{0}^{\pi / 2} \sin 2 x \log \operatorname{ld} x \\
\Rightarrow & \mathrm{I}=0[\because \log 1=0] \\
\therefore & \int_{0}^{\pi / 2} \sin 2 x \log (\tan x) d x=0
\end{aligned}$