Integration using u substitution.?
回答 (3)
2017-02-16 5:46 am
let u =(3/2)x then du=(3/2)dx then dx=du/(3/2)=(2/3)du
int(sin(u) (2/3)du=(2/3)int(sin(u)du=(-)(2/3)cosu+c=(-2/3)cos(3/2)x+c
int(sin(u) (2/3)du=(2/3)int(sin(u)du=(-)(2/3)cosu+c=(-2/3)cos(3/2)x+c
2017-02-16 5:14 am
Let u = 3x
du = 3 dx
du/3 = dx
(1/6) ∫ sin u du = (-1/6) cos u + C
(1/6) ∫ sin u du = (-1/6) cos (3x) + C
du = 3 dx
du/3 = dx
(1/6) ∫ sin u du = (-1/6) cos u + C
(1/6) ∫ sin u du = (-1/6) cos (3x) + C
2017-02-16 2:56 am
- (1/6)cos(3x) + c
收錄日期: 2021-04-24 00:12:37
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20170215184657AAtpPlE