13th Putnam 1953 Problem A4 Using sin x = 2 sin x/2 cos x/2 or otherwise, find ∫ 0 π/2 ln sin x dx.
Let I = ∫ 0 π/2 ln sin x dx. Making the suggested substitution, we get I = π/2 ln 2 + ∫ 0 π/2 ln sin x/2 cos x/2 dx. Putting y = x/2 we get ∫ 0 π/2 ln sin x/2 cos x/2 dx = 2 ∫ 0 π/4 ln sin y + ln cos y dy. But cos y = sin(π/2 - y), so ∫ 0 π/4 ln cos y dy = ∫ π/4 π/2 ln sin y dy. Hence I = π/2 ln 2 + 2I, so I = - π/2 ln 2. 13th Putnam 1953 © John Scholes jscholes@kalva.demon.co.uk 5 Mar 2002