{dx/(x^2+1)^(1/2)
Let x= tant (-兀 /2<兀 2)
dx=(1/(cost)^2)dt
therefore {dx/(x^2+1)^(1/2)
={[1/((tant)^2+1)^(1/2)(1/(cost)^2]dt
={(1/cost)dt
=?
最尾2部唔肯定 高手help!plz!
更新1:
有無人可以唔用sec? 只用sin cos tan?
indefintite integrals
2012-02-29 6:21 am
Find the following indefinite integrals using Integration by Substitution
{dx/(x^2+1)^(1/2)
Let x= tant (-兀 /2<兀 2)
dx=(1/(cost)^2)dt
therefore {dx/(x^2+1)^(1/2)
={[1/((tant)^2+1)^(1/2)(1/(cost)^2]dt
={(1/cost)dt
=?
最尾2部唔肯定 高手help!plz!
{dx/(x^2+1)^(1/2)
Let x= tant (-兀 /2<兀 2)
dx=(1/(cost)^2)dt
therefore {dx/(x^2+1)^(1/2)
={[1/((tant)^2+1)^(1/2)(1/(cost)^2]dt
={(1/cost)dt
=?
最尾2部唔肯定 高手help!plz!
回答 (2)
2012-02-29 6:42 am
✔ 最佳答案
{ (1/cost) dt={ sect dt
={ (sect ( (sect+tant) / (sect+tant) ) dt
={ ( ( (sect)^2+ sect tant ) / (sect+tant) ) )dt
note that ( (sect)^2 + sect tant )dt = d( sect+tant )
then, {(1/cost)dt= { (1/(sect+tant))d(sect+tant)
=ln| sect + tant | + C
=ln| sec(arctanx) + tan(arctanx) | + C
=ln| sec(arctanx) + x | + C
參考: by myself
2012-02-29 5:47 pm
收錄日期: 2021-04-23 18:31:31
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20120228000051KK00774