设f（x）在【0,1】上连续.证明∫（π/2~0)f(cosx)dx=∫（π/2~0)f（sinx）dx

设f（x）在【0,1】上连续.证明∫（π/2~0)f(cosx)dx=∫（π/2~0)f（sinx）dx
设f（x）在【0,1】上连续.证明∫（π/2~0)f(cosx)dx=∫（π/2~0)f（sinx）dx

设f（x）在【0,1】上连续.证明∫（π/2~0)f(cosx)dx=∫（π/2~0)f（sinx）dx
令 y=π/2-x,则x=π/2-y
∫（π/2~0)f(cosx)dx=∫（0~π/2) f(cos(π/2-y))d(π/2-y)
=∫（0~π/2) -f(siny)dy
=-∫（0~π/2) f(siny)dy
=∫（π/2~0)f(siny)dy
=∫（π/2~0)f(sinx)dx

分别证明等号两边的极限值相等。