证明∫(0,π)f(sinx)dx=2∫(0,π/2)f(sinx)dx
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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
(0,π)中,0是下限,π是上限,后面同理,求详解
左边=-cosπ+cos0=2 右边=2(-cosπ/2+cos0)=2 原式成立
再问: 是f(sinx),不是sinx
再答: 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
再问: 是f(sinx),不是sinx
再答: 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
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