令u=π/2 -x
則x=π/2 -u
原積分=∫(π/2→0) f( sin(π/2 -u) ) / [f( sin(π/2 -u) ) + f( cos(π/2 -u) )] d(π/2 -u)
= - ∫(π/2→0) f(cosu) / [f(cosu) + f(sinu)] du
= ∫(0→π/2) f(cosu) / [f(cosu) + f(sinu)] du
積分變量的字符對積分的結果沒有影響,因此
∫(0→π/2)f(sinx)/[f(sinx)+f(cosx)]*dx = ∫(0→π/2)f(cosx)/[f(sinx)+f(cosx)]*dx
則
2 ∫(0→π/2)f(sinx)/[f(sinx)+f(cosx)]*dx
=∫(0→π/2)f(sinx)/[f(sinx)+f(cosx)]*dx + ∫(0→π/2)f(cosx)/[f(sinx)+f(cosx)]*dx
=∫(0→π/2) [f(sinx)+f(cosx)] / [f(sinx)+f(cosx)] dx
=∫(0→π/2) 1 dx
= x|(0→π/2)
= π/2 - 0
= π/2
則原積分 = π/4