设f(x)为连续函数,证明: ∫0πxf(sinx)dx=π/2∫0πf(sinx)dx=π∫0π/2f(sinx)dx;

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问题 设f(x)为连续函数,证明:
0πxf(sinx)dx=π/2∫0πf(sinx)dx=π∫0π/2f(sinx)dx;

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答案令I=∫0πxf(sinx)dx,则I=∫0πxf(sinx)dx→∫π0(π-t)f(sint)(-dt)=∫0π(π-t)f(sint)dt=∫0π(π-x)f(sinx)dx=π∫0πf(sinx)dx-∫0πxf(sinx)dx=π∫0πf(sinx)dx-I,则I=∫0πxf(sinx)dx=π/2 ∫0πf(sinx)dx=π∫0πf(sinx)dx.

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