fx在零到派上连续f0=0到派零到派fx乘以可3xx证明
证明：令t=π-x,则x∈[0,π]时,t∈[π,0] dx=-dt
则I=∫(0→π) xf(sinx) dx
=-∫(π→0) (π-t)f(sin(π-t)) dt
=-∫(π→0) (π-t)f(sint) dt
=∫(0→π)(π-t)f(sint) dt
=∫(0→π)πf(sint) dt-∫(0→π)tf(sint)dt
=∫(0→π)πf(sinx) dx-I
∴I=(1/2)∫(0→π)πf(sinx) dx=π∫(0→π/2)f(sinx) dx
证毕
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