13th Putnam 1953

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Problem A4

Using sin x = 2 sin x/2 cos x/2 or otherwise, find ∫0π/2 ln sin x dx.

 

Solution

Let I = ∫0π/2 ln sin x dx. Making the suggested substitution, we get I = π/2 ln 2 + ∫0π/2 ln sin x/2 cos x/2 dx. Putting y = x/2 we get ∫0π/2 ln sin x/2 cos x/2 dx = 2 ∫0π/4 ln sin y + ln cos y dy. But cos y = sin(π/2 - y), so ∫0π/4 ln cos y dy = ∫π/4π/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