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參考答案:f(cos^2 x)/[f(cos^2 x)+f(sin^2 x)]=1-f(sin^2 x)/[f(cos^2 x)+f(sin^2 x)]
∴∫{0到π/2}f(cos^2 x)/[f(cos^2 x)+f(sin^2 x)]dx=∫{0到π/2}1-∫{0到π/2}f(sin^2 x)/[f(cos^2 x)+f(sin^2 x)]dx
令y=π/2-x，有
∫{0到π/2}f(sin^2 x)/[f(cos^2 x)+f(sin^2 x)]dx
=∫{0到π/2}f(sin^2 y)/[f(cos^2 y)+f(sin^2 y)]dy
=∫{π/2到0}f(sin^2 π/2-x)/[f(cos^2 π/2-x)+f(sin^2 π/2-x)]d(π/2-x)
=∫{0到π/2}f(cos^2 x)/[f(sin^2 x)+f(cos^2 x)]dx
注意中间d(π/2-x)变到dx和{π/2到0}变到{0到π/2}两次改变符号
∴∫{0到π/2}f(cos^2 x)/[f(cos^2 x)+f(sin^2 x)]dx=(∫{0到π/2}1)/2=π/4