Question

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

Answer 1

左边=-cosπ+cos0=2 右边=2(-cosπ/2+cos0)=2 原式成立

追问

是f(sinx),不是sinx

追答

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