integrate e^(-3x) tan(e^(-3x)) dx?
回答 (2)
2016-09-10 2:38 am
u = cos(e^(−3x))
du = −sin(e^(−3x)) * e^(−3x) * −3 dx
du = 3e^(−3x) sin(e^(−3x)) dx
∫ e^(−3x) tan(e^(−3x)) dx
= 1/3 ∫ 3e^(−3x) sin(e^(−3x)) / cos(e^(−3x)) dx
= 1/3 ∫ du/u
= 1/3 ln|u| + C
= 1/3 ln|cos(e^(−3x))| + C
2016-09-10 2:42 am
The e^(-3x) factor suggests u substitution.
let u = e^(-3x)
du = -3e^(-3x)
∫ e^(-3x) tan(e^(-3x)) dx
= -1/3 ∫ tan(u) du
= (-1/3) (-ln|cos(u)|)
= 1/3 ln|cos(e^(-3x))| + C
let u = e^(-3x)
du = -3e^(-3x)
∫ e^(-3x) tan(e^(-3x)) dx
= -1/3 ∫ tan(u) du
= (-1/3) (-ln|cos(u)|)
= 1/3 ln|cos(e^(-3x))| + C
收錄日期: 2021-04-21 21:43:26
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160909182910AAEcc22