rjilkf 發問時間： 科學數學 · 6 年前

# 微積分∫dy=∫4e−xcos(x)dx(急!!!

∫dy=∫4e−xcos(x)dx

∫dy=uv−∫vdu關於這個算法我有看不懂阿....

∫dy=∫4e^−x cos(x)dx -x是次方

### 2 個解答

• KlIE
Lv 6
6 年前
最佳解答

−xcos(x)整個都是exp 指數函數的次方嗎 ?

假如只有-x是次方就簡單多了...

2014-09-30 14:06:49 補充：

假如 只有−x 是exp 指數函數的次方 ( 若−xcos(x) 整個都是次方 .難度太高 ,不會...)

那運算過程如下

∫dy=∫ 4 exp(-x) cos(x)dx = 4 *∫ exp(-x) cos(x)dx ..... (a式)

令u = exp (-x) , dv = cos(x) dx

du = exp (-x) (-x)' dx = -exp (-x) dx , v =sin (x)

∫ exp(-x) cos(x)dx

= ∫ u dv

= uv -∫ vdu

=[exp (-x)] [ sin (x)] -∫ [sin (x)] [-exp (-x)] dx

=[exp (-x)] [ sin (x)] +∫ [sin (x)] [exp (-x)] dx .... (b式)

欲求 ∫ exp(-x) cos(x)dx ,需先求 ∫ [sin (x)] [exp (-x)] dx = ?

令 u' = exp (-x) , dv'= sin(x) dx

du' = -exp (-x) dx , v'= -cos (x)

==> ∫ [sin (x)] [exp (-x)] dx

= ∫ [exp (-x)] [sin (x)] dx

= ∫ u'dv'

= u'v' -∫ v'du'

= [exp (-x)][-cos (x)] - ∫ [-cos (x)][- exp (-x) dx]

= -[exp (-x)] [cos (x)] - ∫ [cos (x)][ exp (-x)] dx ... (c式)

c式帶回b式

∫ exp(-x) cos(x)dx

=[exp (-x)] [ sin (x)] +∫ [sin (x)] [exp (-x)] dx

= [exp (-x)] [ sin (x)] + {-[exp (-x)] [cos (x)] - ∫ [cos (x)][ exp (-x)] dx }

移項

=> 2* ∫ exp(-x) cos(x)dx = exp (-x)] [ sin (x)] -[exp (-x)] [cos (x)] + C1 ...(d式)

(C1 : 積分常數)

d式帶回a式

∫dy=∫ 4 exp(-x) cos(x)dx = 4 *∫ exp(-x) cos(x)dx

= 2*{[exp (-x)] [ sin (x)] -[exp (-x)] [cos (x)]} +2C1

=2 exp(-x) [ sin (x) - cos (x) ] + C (C : 積分常數)

得解#

2014-09-30 14:10:18 補充：

第二次部分積分時

若令 u' = sin(x), dv'= exp (-x) dx

所得c式結果 ,帶回 b式

會發現等號兩邊消光光

有興趣可以試試看

參考資料： Calculus with Analytic Geometry
• 6 年前

對,是次方~~~~~