Evaluate $\int\limits_{0}^{\frac{\pi}{2}} \log \sin x \, dx$

$-\frac{\pi}{2} \log 2$

The correct answer is Option (2) → $-\frac{\pi}{2} \log 2$

Let $I = \int\limits_{0}^{\frac{\pi}{2}} \log \sin x \, dx$. Then, by $P_4$

$I = \int\limits_{0}^{\frac{\pi}{2}} \log \sin \left( \frac{\pi}{2} - x \right) dx = \int_{0}^{\frac{\pi}{2}} \log \cos x \, dx$

Adding the two values of $I$, we get

$2I = \int\limits_{0}^{\frac{\pi}{2}} (\log \sin x + \log \cos x) \, dx$

$= \int\limits_{0}^{\frac{\pi}{2}} (\log \sin x \cos x + \log 2 - \log 2) \, dx \text{ (by adding and subtracting } \log 2)$

$= \int\limits_{0}^{\frac{\pi}{2}} \log \sin 2x \, dx - \int_{0}^{\frac{\pi}{2}} \log 2 \, dx \quad \text{(Why?)}$

Put $2x = t$ in the first integral. Then $2 \, dx = dt$, when $x = 0, t = 0$ and when $x = \frac{\pi}{2}, t = \pi$.

Therefore,

$2I = \frac{1}{2} \int\limits_{0}^{\pi} \log \sin t \, dt - \frac{\pi}{2} \log 2$

$= \frac{2}{2} \int\limits_{0}^{\frac{\pi}{2}} \log \sin t \, dt - \frac{\pi}{2} \log 2 \quad [\text{by } P_6 \text{ as } \sin(\pi - t) = \sin t]$

$= \int\limits_{0}^{\frac{\pi}{2}} \log \sin x \, dx - \frac{\pi}{2} \log 2 \quad \text{(by changing variable } t \text{ to } x)$

$= I - \frac{\pi}{2} \log 2$

Hence, $\int\limits_{0}^{\frac{\pi}{2}} \log \sin x \, dx = \frac{-\pi}{2} \log 2$.