我的做法比較繁瑣，不知是否有人可以提供更簡單的方法，
我的做法如下：
\(\displaystyle\int^{\pi/2}_0\ln\left(\sin x\right)dx = \int^{\pi/2}_0\ln\left(\cos \left(\frac{\pi}{2}-x\right)\right)dx \)
\(\displaystyle=\int^0_{\pi/2}\ln\left(\cos x\right)\left(-dx\right) = \int^{\pi/2}_0\ln\left(\cos x\right)dx \)
且
\(\displaystyle\int^{\pi/2}_0\ln\left(\sin x\right)dx = \int^{\pi/2}_0\ln\left(\sin \left(\pi-x\right)\right)dx \)
\(\displaystyle=\int^{\pi/2}_{\pi}\ln\left(\sin x\right)\left(-dx\right) = \int^{\pi}_{\pi/2}\ln\left(\sin x\right)dx \)
然後，利用上面兩式，可得如下
\(\displaystyle2\int^{\pi/2}_0\ln\left(\sin\left(x\right)\right)dx\)
\(\displaystyle=\int^{\pi/2}_0\ln\left(\sin x\right)dx +\int^{\pi/2}_0\ln\left(\cos x\right)dx \)
\(\displaystyle=\int^{\pi/2}_0\left(\ln\left(\sin x\right)+\ln\left(\cos x\right)\right)dx \)
\(\displaystyle=\int^{\pi/2}_0\ln\left(\frac{\sin\left(2x\right)}{2}\right)dx \)
\(\displaystyle=\int^{\pi/2}_0\ln\left(\sin\left(2x\right)\right)dx -\int^{\pi/2}_0\ln2 dx\)
\(\displaystyle=\int^{\pi}_0\ln\left(\sin\left(x\right)\right)\frac{dx}{2} -\int^{\pi/2}_0\ln2 dx\)
\(\displaystyle=\frac{1}{2}\left(\int^{\pi/2}_0\ln\left(\sin\left(x\right)\right)dx+\int^{\pi}_{\pi/2}\ln\left(\sin\left(x\right)\right)dx\right) -\int^{\pi/2}_0\ln2 dx\)
\(\displaystyle=\int^{\pi/2}_0\ln\left(\sin\left(x\right)\right)dx -\int^{\pi/2}_0\ln2 dx\)
因此，
\(\displaystyle\int^{\pi/2}_0\ln\left(\sin\left(x\right)\right)dx = -\int^{\pi/2}_0\ln2 dx=-\frac{\pi}{2}\ln2.\)
題目出處：99台中一中教甄
