Problem Statement
Evaluate: $I = \int_{0}^{\pi/2} \log(\sin x) \, dx$.
Verified Solution & Marking Scheme
Apply King Property
$I = \int_{0}^{\pi/2} \log(\cos x) \, dx$.
Add Equations
$$2I = \int_{0}^{\pi/2} [\log(\sin x) + \log(\cos x)] \, dx = \int_{0}^{\pi/2} \log(\sin 2x) \, dx - \int_{0}^{\pi/2} \log 2 \, dx$$
Substitute 2x = t
$$\int_{0}^{\pi/2} \log(\sin 2x) \, dx = \frac{1}{2} \int_{0}^{\pi} \log(\sin t) \, dt = \int_{0}^{\pi/2} \log(\sin t) \, dt = I$$
$$2I = I - \frac{\pi}{2} \log 2 \implies I = -\frac{\pi}{2} \log 2$$