∫f（sinx,cosx）dx=∫f（cosx,sinx）dx上下限是[0,π/2]
问题描述：
∫f（sinx,cosx）dx=∫f（cosx,sinx）dx上下限是[0,π/2]
x=π/2-t 我不明白为什么要假设 而且sinx和cosx在π/2之间的转化不清楚
答
补充楼上的回答∫[0,π/2]f(sinx,cosx)dx x=π/2-u x=0,u=π/2,x=π/2,u=0=∫[π/2,0] f(sin(π/2-u),cos(π/2-u))d(π/2-u)=-∫[π/2,0]f(cosu,sinu)du=∫[0,π/2]f(cosu,sinu)du∫[0,π/2]f(sinx)dx=∫[0,π/2]f(c...