1.∫3-cot^2x/co^2x dx
2.∫1+cos2x/1-cos2x dx
不定積分問題x2........(十分緊急
2006-12-11 6:33 am
回答 (1)
2006-12-11 6:40 am
✔ 最佳答案
∫3-cot^2x/cot^2x dx=∫(3sec^2x-1)dx
=3tanx -x +C
∫3-cot^2x/cos^2x dx
=∫(3sec^2x-csc^2x)dx
=3tanx +cotx +C
∫(1+cos2x)/(1-cos2x) dx
=∫(2cos^2x)/(2sin^2x) dx
=∫cot^2x dx
=∫(csc^2x -1)dx
=-cotx -x +C
2006-12-11 00:42:36 補充：
更正∫3-cot^2x/cot^2x dx=∫(3tan^2x-1)dx=∫(3sec^2x-4)dx=3tanx -4x C
收錄日期: 2021-04-12 21:22:52
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20061210000051KK05602