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

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