证明∫(0,π)f(sinx)dx=2∫(0,π/2)f(sinx)dx
问题描述：
证明∫(0,π)f(sinx)dx=2∫(0,π/2)f(sinx)dx
(0,π)中,0是下限,π是上限,后面同理,求详解
答
左边=-cosπ+cos0=2 右边=2(-cosπ/2+cos0)=2 原式成立是f(sinx),不是sinx抱歉，没仔细看题呵。令x=(π/2)-t 则∫(0,π/2)f(sinx)dx=∫(π/2,0)f(cost)(-dt)=∫(0,π/2)f(cost)dt∫(0,π)f(sinx)dx=∫(π/2,-π/2)f(cost)(-dt)=∫(-π/2,π/2)f(cost)dt 又f(cost)为偶函数，故∫(-π/2,π/2)f(cost)dt=∫(-π/2,0)f(cost)dt+∫(0,π/2)f(cost)dt=2∫(0,π/2)f(cost)dt故∫(0,π)f(sinx)dx=2∫(0,π/2)f(sinx)dx