Question:

<div class='solution'> <p>Let $I = \int_0^{\pi/2}\ln\sin x\,dx$. King: $I = \int_0^{\pi/2}\ln\cos x\,dx$ (replace $x\to\pi/2-x$).</p> <p>Add: $2I = \int_0^{\pi/2}\ln(\sin x\cos x)\,dx = \int_0^{\pi/2}\ln\frac{\sin 2x}{2}\,dx$</p> <p>$$= \int_0^{\pi/2}\ln\sin 2x\,dx - \frac{\pi}{2}\ln 2$$</p> <p>Sub $t=2x$ in first integral: $\int_0^{\pi/2}\ln\sin 2x\,dx = \frac{1}{2}\int_0^\pi\ln\sin t\,dt = \frac{1}{2}\cdot 2\int_0^{\pi/2}\ln\sin t\,dt = I$</p> <p>(Using symmetry: $\int_0^\pi\ln\sin t\,dt = 2\int_0^{\pi/2}\ln\sin t\,dt$)</p> <p>$$2I = I - \frac{\pi}{2}\ln 2 \Rightarrow I = \boxed{-\frac{\pi}{2}\ln 2}$$</p>