Find ∫0π/2 f(x) dx, where f(x) = 1/(1 + tan√2x).
This involves a trick. f(π/2 - x) = 1/(1 + cot√2x) = tan√2x/(1 + tan√2x) = 1 - f(x). Hence ∫0π/2 f(x) dx = ∫0π/4 f(x) dx + ∫π/4π/2 f(x) dx = ∫0π/4 f(x) dx + ∫0π/4 f(π/2 - x) dx = ∫0π/4 f(x) dx + ∫0π/4 (1 - f(x) ) dx = ∫0π/4 dx = π/4.
Note that the √2 is irrelevant - the argument works for any exponent.
41st Putnam 1980
© John Scholes
10 Dec 1999
Last corrected/updated 21 Nov 02