设f(u)是连续函数,证明:∫0πxf(sinx)dx=π/2∫0πf(sinx)dx,并求∫0π

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问题 设f(u)是连续函数,证明:∫0πxf(sinx)dx=π/2∫0πf(sinx)dx,并求∫0π

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答案I=∫0πxf(sinx)dx[*]614∫π0(π-t)f(sint)(-dt) =π∫0πf(sint)dt-∫0πtf(sint)dt=π∫0πs(sinx)dx-∫0πxf(sinx)dx =π∫0πf(sinx)dt-I, 则∫0πxf(sinx)dx=π/2∫0πf(sinx)dx. [*]

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