Result: I = integral(0 to pi/2) ln(sin x) dx = -(pi/2)*ln 2

Proof: Step 1: By King's Rule, I = integral(0 to pi/2) ln(cos x) dx. So integral(0 to pi/2) ln(sin x) dx = integral(0 to pi/2) ln(cos x) dx.

Step 2: Add: 2I = integral(0 to pi/2) ln(sin x * cos x) dx = integral(0 to pi/2) ln(sin 2x/2) dx = integral(0 to pi/2) ln(sin 2x) dx - (pi/2)*ln 2.

Step 3: Let J = integral(0 to pi/2) ln(sin 2x) dx. Substitute t = 2x: J = (1/2)*integral(0 to pi) ln(sin t) dt.

Step 4: Using Queen's Rule on [0, pi]: integral(0 to pi) ln(sin t) dt = 2*integral(0 to pi/2) ln(sin t) dt = 2I.

Step 5: So J = (1/2)*2I = I. From Step 2: 2I = I - (pi/2)*ln 2. Therefore I = -(pi/2)*ln 2.