∫₀^(π/2) log(sin x) dx equals:

∫₀^(π/2) log(sin x) dx equals:

• A. π log 2
• B. (π/2) log 2
• C. −(π/2) log 2
• D. −π log 2

Answer: C) −(π/2) log 2

Explanation: Let I = ∫₀^(π/2) log(sin x) dx. By property, I = ∫₀^(π/2) log(cos x) dx. Adding: 2I = ∫₀^(π/2) log(sin x cos x) dx = ∫₀^(π/2) log(sin 2x / 2) dx = ∫₀^(π/2) log(sin 2x) dx − ∫₀^(π/2) log 2 dx. Let 2x = t → dx = dt/2. Limits 0 to π. First integral = (1/2) ∫₀^π log(sin t) dt = (1/2) × 2 ∫₀^(π/2) log(sin t) dt = I. So 2I = I − (π/2) log 2 → I = −(π/2) log 2.