※ 引述《young11539 (〝☆小小霈★”)》之铭言： : if f(x) conti. on [0,1] : then show that : ∫xf(sinx)dx = (π/2)∫f(sinx)dx : 积分范围 [0,π] : 感激不尽 设 x=π-y, dx=-dy π ∫ xf(sinx)dx 0 0 = ∫(π-y)f(sin(π-y))(-dy) π π = ∫(π-y)f(siny)dy 0 π π = π∫ f(siny)dy - ∫yf(siny)dy 0 0 ∵变数不影响积分值 π π π ∴∫ xf(sinx)dx = π∫ f(sinx)dx - ∫xf(sinx)dx 0 0 0 π π =>∫ xf(sinx)dx = (π/2)∫ f(sinx)dx # 0 0 --

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