The value of $\int\limits_0^{\pi / 2}(2 \log \sin x-\log \sin 2 x) d x$, is
Answer & explanation
Correct answer: option 2
Let $I=\int\limits_0^{\pi / 2}(2 \log \sin x-\log \sin 2 x) d x$. Then,
$I=\int\limits_0^{\pi / 2}\{2 \log \sin x-\log (2 \sin x \cos x)\} d x$
$\Rightarrow I=\int\limits_0^{\pi / 2}\{2 \log \sin x-\log 2-\log \sin x-\log \cos x\} d x$
$\Rightarrow I=\int\limits_0^{\pi / 2} \log \sin x d x-\int\limits_0^{\pi / 2} \log 2 d x-\int\limits_0^{\pi / 2} \log \cos x d x$
$\Rightarrow I=\int\limits_0^{\pi / 2} \log \sin x d x-(\log 2) \int\limits_0^{\pi / 2} d x-\int\limits_0^{\pi / 2} \log \cos \left(\frac{\pi}{2}-x\right) d x$
$\Rightarrow I=\int\limits_0^{\pi / 2} \log \sin x d x-(\log 2)[x]_0^{\pi / 2}-\int\limits_0^{\pi / 2} \log \sin x d x$
$\Rightarrow I=-(\log 2)\left(\frac{\pi}{2}-0\right)=-\frac{\pi}{2} \log 2$