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

(C) Let $I = \int_{0}^{\frac{\pi}{2}} \log \sin x \, dx$
Using the property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$,we have:
$I = \int_{0}^{\frac{\pi}{2}} \log \sin \left(\frac{\pi}{2}-x\right) \, dx = \int_{0}^{\frac{\pi}{2}} \log \cos x \, dx$
Adding the two expressions for $I$:
$2I = \int_{0}^{\frac{\pi}{2}} (\log \sin x + \log \cos x) \, dx = \int_{0}^{\frac{\pi}{2}} \log(\sin x \cos x) \, dx$
Multiply and divide by $2$ inside the logarithm:
$2I = \int_{0}^{\frac{\pi}{2}} \log\left(\frac{\sin 2x}{2}\right) \, dx = \int_{0}^{\frac{\pi}{2}} \log \sin 2x \, dx - \int_{0}^{\frac{\pi}{2}} \log 2 \, dx$
$2I = \int_{0}^{\frac{\pi}{2}} \log \sin 2x \, dx - \frac{\pi}{2} \log 2$
For the first integral,let $2x = t$,then $2 \, dx = dt$. When $x=0, t=0$; when $x=\frac{\pi}{2}, t=\pi$:
$\int_{0}^{\frac{\pi}{2}} \log \sin 2x \, dx = \frac{1}{2} \int_{0}^{\pi} \log \sin t \, dt = \frac{1}{2} \times 2 \int_{0}^{\frac{\pi}{2}} \log \sin t \, dt = I$ (using $\int_{0}^{2a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$ if $f(2a-x) = f(x)$)
Substituting back:
$2I = I - \frac{\pi}{2} \log 2$
$I = -\frac{\pi}{2} \log 2$