求e^2xcos3x對x的積分
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求e^2xcos3x對x的積分(急件)
2009-12-12 4:24 am
回答 (3)
2009-12-12 4:37 am
✔ 最佳答案
Please refer to the followings:http://i726.photobucket.com/albums/ww265/physicsworld2010/physicsworld01Dec112036.jpg?t=1260535011
圖片參考：http://i726.photobucket.com/albums/ww265/physicsworld2010/physicsworld01Dec112036.jpg?t=1260535011
參考: Physics king
2009-12-12 11:01 am
用特殊密計算比較快~~
2009-12-12 5:38 am
1. S f g’= fg- S f’g
原式=(1/3) S e^2x(cos3x)3=
(1/3) [ e^2x(sin3x)- S (e^2x)(2) (sin3x)]
=(1/3) e^2x(sin3x)-2/3 S (e^2x) (sin3x)
2. S f g’= fg- S f’g
S (e^2x) (sin3x)
=[-1/3] S e^2x(-sin3x)3
= [-1/3]( (e^2x)cos3x- S (e^2x)(2) cos3x)
=[-1/3] (e^2x)cos3x+(2/3) S (e^2x) cos3x
把2.代入1.
3.
S e^2x(cos3x)
= (1/3) e^2x(sin3x)+2/9 e^2x(cos3x)-4/9 S e^2x(cos3x)
整理得:
13/9 S e^2x(cos3x)=(1/3) e^2x(sin3x)+(2/9) e^2x(cos3x)
S e^2x(cos3x)=(3/13) e^2x(sin3x)+(2/13) e^2x(cos3x)
原式=(1/3) S e^2x(cos3x)3=
(1/3) [ e^2x(sin3x)- S (e^2x)(2) (sin3x)]
=(1/3) e^2x(sin3x)-2/3 S (e^2x) (sin3x)
2. S f g’= fg- S f’g
S (e^2x) (sin3x)
=[-1/3] S e^2x(-sin3x)3
= [-1/3]( (e^2x)cos3x- S (e^2x)(2) cos3x)
=[-1/3] (e^2x)cos3x+(2/3) S (e^2x) cos3x
把2.代入1.
3.
S e^2x(cos3x)
= (1/3) e^2x(sin3x)+2/9 e^2x(cos3x)-4/9 S e^2x(cos3x)
整理得:
13/9 S e^2x(cos3x)=(1/3) e^2x(sin3x)+(2/9) e^2x(cos3x)
S e^2x(cos3x)=(3/13) e^2x(sin3x)+(2/13) e^2x(cos3x)
收錄日期: 2021-04-19 20:48:36
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20091211000016KK06404