indefintite integrals

👨🏻‍🏫 學術台數學

郁林

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!

更新1:

有無人可以唔用sec? 只用sin cos tan?

回答 (2)

JHO

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

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sir

2012-02-29 5:47 pm

避開sec 的方法:

圖片參考：http://imgcld.yimg.com/8/n/HA00132885/o/701202280077413873442830.jpg

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收錄日期: 2021-04-23 18:31:31
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