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xe^-x^2cote^-x^2dx的不定积分
人气:180 ℃ 时间:2020-03-20 20:49:38
解答
∫xe^(-x^2)cote^(-x^2)dx
= -(1/2)∫e^(-x^2)cote^(-x^2)d(-x^2)
= -(1/2)∫cose^(-x^2)/sin[e^(-x^2)]d[e^(-x^2)]
= -(1/2)∫1/sin[e^(-x^2)]d[sine^(-x^2)]
= -(1/2)ln|sine^(-x^2)|+C
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