Practicing Success
The value of $\int\limits_0^{\pi / 2}(2 \log \sin x-\log \sin 2 x) d x$, is |
$\frac{\pi}{2} \log 2$ $-\frac{\pi}{2} \log 2$ $\pi \log 2$ $-\pi \log 2$ |
$-\frac{\pi}{2} \log 2$ |
Let $I=\int\limits_0^{\pi / 2}(2 \log \sin x-\log \sin 2 x) d x$. Then, $I=\int\limits_0^{\pi / 2}\{2 \log \sin x-\log (2 \sin x \cos x)\} d x$ $\Rightarrow I=\int\limits_0^{\pi / 2}\{2 \log \sin x-\log 2-\log \sin x-\log \cos x\} d x$ $\Rightarrow I=\int\limits_0^{\pi / 2} \log \sin x d x-\int\limits_0^{\pi / 2} \log 2 d x-\int\limits_0^{\pi / 2} \log \cos x d x$ $\Rightarrow I=\int\limits_0^{\pi / 2} \log \sin x d x-(\log 2) \int\limits_0^{\pi / 2} d x-\int\limits_0^{\pi / 2} \log \cos \left(\frac{\pi}{2}-x\right) d x$ $\Rightarrow I=\int\limits_0^{\pi / 2} \log \sin x d x-(\log 2)[x]_0^{\pi / 2}-\int\limits_0^{\pi / 2} \log \sin x d x$ $\Rightarrow I=-(\log 2)\left(\frac{\pi}{2}-0\right)=-\frac{\pi}{2} \log 2$ |