THE WILLIAM LOWELL PUTNAM MATHEMATICAL COMPETITION.

and then note that sin(θ)+cos(θ) = √ 2 cos(π/4− θ). We may thus rewrite the integrand as 1 2 log(2) + log(cos(π/4− θ))− log(cos(θ)). But over the interval [0, π/4], the integrals of log(cos(θ)) and log(cos(π/4−θ)) are equal, so their contributions cancel out. The desired integral is then just the integral of 12 log(2) over the interval [0, π/4], which is π log(2)/8. Second solution: (by Roger Nelsen) Let I denote the desired integral. We make the substitution x = (1− u)/(1 + u) to obtain