求 4 cosx / 1 - (cosx)^2 的積分,,,,
連埋詳細的STEP
[微積分]積分問題
2007-04-14 1:54 am
回答 (2)
2007-04-14 4:23 am
✔ 最佳答案
除了樓上朋友的方法，你不妨參考以下的方法，會比較直接些∫[4cosx/(1-(cosx)^2)]dx
=∫[4cosx/(sinx)^2]dx
=∫4dsinx/(sinx)^2
= - 4/sinx + C
= -4 cscx + C
2007-04-14 2:10 am
打唔到積分符號,用Ｓ頂住先
Ｓ { 4cosx / [1-(cos x)^2] } dx
= Ｓ (4cosx / sin^2 x) dx
= Ｓ [4(1/sinx)(cosx/sinx)] dx
= Ｓ (4cscx cotx) dx
= - 4 csc x + C , where C is a constant. [d/dx (csc x) = - csc x cot x]
Ｓ { 4cosx / [1-(cos x)^2] } dx
= Ｓ (4cosx / sin^2 x) dx
= Ｓ [4(1/sinx)(cosx/sinx)] dx
= Ｓ (4cscx cotx) dx
= - 4 csc x + C , where C is a constant. [d/dx (csc x) = - csc x cot x]
參考: by My Maths
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