What is ∫₀^(π/2) (a cosx + b sinx)/(sinx + cosx) dx equal to?

From the original integral, ∫₀^(π/2) (a sinx + b cosx)/[(a+b)(sinx+cosx)] dx = k. So ∫₀^(π/2) (a sinx + b cosx)/(sinx+cosx) dx = k(a+b). By the symmetry property used in Q83, ∫₀^(π/2) (a cosx + b sinx)/(sinx+cosx) dx also equals k(a+b) (since both equal π/2·(a+b)/2 = (a+b)π/4 = k(a+b)).