xe^-x^2cote^-x^2dx的不定积分

1个回答

∫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