rjilkf asked in 科學數學 · 5 years ago

微積分∫dy=∫4e−xcos(x)dx(急!!!

我想請問一下

∫dy=∫4e−xcos(x)dx

這一題該怎麼算??

∫dy=uv−∫vdu關於這個算法我有看不懂阿....

請求高人教導一下謝謝.....

Update:

∫dy=∫4e^−x cos(x)dx -x是次方

2 Answers

Rating
  • KlIE
    Lv 6
    5 years ago
    Favorite Answer

    −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

    Source(s): Calculus with Analytic Geometry
  • 5 years ago

    對,是次方~~~~~

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