Statement - 1: $\int\limits_0^{\pi / 2} x \cot x d x=\frac{\pi}{2} \log 2$
Statement - 2: $\int\limits_0^{\pi / 2} \log \sin x d x=-\frac{\pi}{2} \log 2$
Answer & explanation
Correct answer: option 1
Clearly, statement-2 is true.
Now, $\int\limits_0^{\pi / 2} x \cot x d x=[x \log \sin x]_0^{\pi / 2}-\int\limits_0^{\pi / 2} \log \sin x d x$
$\Rightarrow \int\limits_0^{\pi / 2} x \cot x d x=0-\lim _{x \rightarrow 0} x \log \sin x-\left(-\frac{\pi}{2} \log 2\right)$
$\Rightarrow \int\limits_0^{\pi / 2} x \cot x d x=\frac{\pi}{2} \log 2$
So, statement-1 is true. Also, Statement-2 is a correct explanation for statement.