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若f(x)在[0,1]上连续,证明 ∫【上π/2下0】f(sinx)dx= ∫【上π/2下0】f(cosx)dx
人气:224 ℃ 时间:2019-08-20 12:15:23
解答
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
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