不定積分問題x2........(十分緊急

2006-12-11 6:33 am
1.∫3-cot^2x/co^2x dx

2.∫1+cos2x/1-cos2x dx

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
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