Result: I = integral$\frac{0 to pi}{2}$ ln(sin x) dx = -$\frac{pi}{2}$*ln 2

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

Step 2: Add: 2I = integral$\frac{0 to pi}{2}$ ln(sin x * cos x) dx = integral$\frac{0 to pi}{2}$ ln$\frac{sin 2x}{2}$ dx = integral$\frac{0 to pi}{2}$ ln(sin 2x) dx - $\frac{pi}{2}$*ln 2.

Step 3: Let J = integral$\frac{0 to pi}{2}$ ln(sin 2x) dx. Substitute t = 2x: J = $\frac{1}{2}$*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$\frac{0 to pi}{2}$ ln(sin t) dt = 2I.

Step 5: So J = $\frac{1}{2}$*2I = I. From Step 2: 2I = I - $\frac{pi}{2}$*ln 2. Therefore I = -$\frac{pi}{2}$*ln 2.