设函数f(x)在[-π,π]上连续、恒正(π是Pai),且f(x)f(-x)=1,则∫(上限π,下限-π)(cosx)^

∫[-π→π] cos²x/[1+f(x)] dx
令x=-u,则u：π→-π,dx=-du
=-∫[π→-π] cos²u/[1+f(-u)] du
先交换上下限
=∫[-π→π] cos²u/[1+f(-u)] du
分子分母同乘以f(u)
=∫[-π→π] cos²uf(u)/[f(u)+f(-u)f(u)] du
=∫[-π→π] cos²uf(u)/[f(u)+1] du
将积分变量换成x
=∫[-π→π] cos²xf(x)/[f(x)+1] du
左边=右边,因此左边=(1/2)(左边+右边)
∫[-π→π] cos²x/[1+f(x)] dx
=(1/2){∫[-π→π] cos²x/[1+f(x)] dx + ∫[-π→π] cos²xf(x)/[f(x)+1] du }
=(1/2)∫[-π→π] cos²x[1+f(x)]/[1+f(x)] dx
=(1/2)∫[-π→π] cos²x dx
=(1/4)∫[-π→π] (1+cos2x) dx
=π/2
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