∫f（sinx,cosx）dx=∫f（cosx,sinx）dx上下限是[0,π/2]

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(cos(π/2-x)dx π/2-x=u
=∫[π/2,0]f(cosu)d(-u)
=∫[0,π/2]f(cosu)du
观察y=sinx和y=cosx图像
y=sinx=cos(x-π/2)=cos(π/2-x)=cosx
(1/2)(π/2-x+x)=π/4
可以发现sinx和cosx关于x=π/4对称,
因此在[0,π/2]积分区间内,设x=π/2-t