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∫(sinx)^2/[1+e^(-x)] dx 积分上下限(π/4,π/4)怎么算?
人气:191 ℃ 时间:2020-03-26 04:49:43
解答
注:此题的上下限有错,应该是积分上下限(-π/4,π/4)!原式=∫(-π/4,π/4)(sinx)^2/[1+e^(-x)]dx (∫(-π/4,π/4)表示从-π/4到π/4积分)=∫(-π/4,0)(sinx)^2/[1+e^(-x)]dx+∫(0,π/4)(sinx)^2/[1+e^(-x)]dx=-∫(π...
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