1/2
∫ (1 - x^2)^ (-1/2) dx
0
[hint : let x = sinθ ]
請問點in 呀??? 我 in 數真的很差....m(__)m
有關 amath ce 02 年的.....
2007-04-08 12:00 am
回答 (2)
2007-04-08 12:12 am
✔ 最佳答案
putx=sinθ,dx=cosθdθwhen x=1/2,θ=@/6 (@=pie)
x=0,θ=0
1/2
∫ (1 - x^2)^ (-1/2) dx
0
= @/6
∫ (1-sin^2 θ)^ (-1/2) cosθdθ
0
= 1/2
∫ (cosθ/cosθ)dθ
0
= @/6
∫ dθ
0
= @/6
[θ]
0
=@/6
good
參考: ME
2007-04-08 7:37 am
it is out c la bo~
收錄日期: 2021-04-25 14:51:17
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070407000051KK02634