佑信 asked in 科學數學 · 9 years ago

Calculus - integral(cos(sinx))

integral cosx(cos(sinx))^5 dx=?

請詳述過程!

Update:

=integral cos(y)^4*d(siny)

=integral [1-sin(y)^2]^4*d(siny)

次方數好像怪怪的?

4 Answers

Rating
  • 金鼎
    Lv 5
    9 years ago
    Favorite Answer

    大大你好,由我為你解答

    ∫cosx*(cos(sinx))^5 dx

    (這題需要帶換2次,答案才會出來)

    Sol: u=sinx

    du=cosxdx

    =∫cos^5(u) du

    =∫cos^4(u)*cos(u) du

    v=sinu

    dv=cosudu

    =∫(1-sin^2)^2 dv

    =∫(1-v^2)^2 dv

    =∫(1-2v^2+v^4)dv

    = v-2/3v^3+1/5v^5+c

    (再來把值帶入還原)

    =sinu-2/3sinu^3+1/5sinu^5+c

    = sin(sinx)-2/3sin(sinx)^3+1/5sin(sinx)^5+c為正解

    Source(s):
  • 麻辣
    Lv 7
    9 years ago

    a=∫cosx*(cos(sinx))^5 dx=?=∫[cos(sinx)]^5*d(sinx)=∫cos(y)^5*dy..........y=sin(x)=∫cos(y)^4*d(siny)=∫[1-sin(y)^2]^4*d(siny)=∫(1-u^2)^4*du.........u=sin(y)=∫(1-4u^2+6u^4-4u^6+u^8)du=u-4u^3/3+6u^5/5-4u^7/7+u^9/9+c=sin(y)-4/3*sin(y)^3+6/5*sin(y)^5-4/7*sin(y)^7+1/9*sin(y)^9+c=sin(sinx)-4/3*sin(sinx)^3+6/5*sin(sinx)^5-4/7*sin(sinx)^7+1/9*sin(sinx)^9+c

  • 9 years ago

    答案出來有sin(sinx)

    ...

    這題目怎麼怪怪的...

  • 9 years ago

    變數代換, 令 u=sin(x).

    然後 [cos(u)]^5 = [1-(sin(u))^2]^2 cos(u) 展開.

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