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Math Help - Integration with parts repeatedly

  1. #1
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    Integration with parts repeatedly

    \int x^5cos(x^3)dx

    dv = cos(x^3)
    u= x^5
    v = sin x^3 / 3
    du = 5x^4

    With those settings I can see that I'm going to have to integrate by parts several times, and by several I mean more than 4 times. Is there a faster way to solve this problem that I don't realize?
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  2. #2
    Member Miss's Avatar
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    Quote Originally Posted by Archduke01 View Post
    \int x^5cos(x^3)dx

    dv = cos(x^3)
    u= x^5
    v = sin x^3 / 3
    du = 5x^4

    With those settings I can see that I'm going to have to integrate by parts several times, and by several I mean more than 4 times. Is there a faster way to solve this problem that I don't realize?
    Your v is wrong.
    You can not find v in terms of the elementary functions.
    You need the gamma function. Which is ,surely, you did not cover it yet.

    If I were you, I will rewrite x^5 as x^3 \, x^2 then I will use the substitution t=x^3 then I will integrate it by parts
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  3. #3
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    Quote Originally Posted by Archduke01 View Post
    \int x^5cos(x^3)dx

    dv = cos(x^3)
    u= x^5
    v = sin x^3 / 3
    du = 5x^4

    With those settings I can see that I'm going to have to integrate by parts several times, and by several I mean more than 4 times. Is there a faster way to solve this problem that I don't realize?
    If you use u=x^3

    then \frac{du}{dx}=3x^2\ \Rightarrow\ du=3x^2dx

    \frac{1}{3}udu=x^5dx

    the integral becomes

    \frac{1}{3}\int{ucos(u)}du
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  4. #4
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    Quote Originally Posted by Miss View Post
    Your v is wrong.
    You can not find v in terms of the elementary functions
    But, then what is my v if my dv is cos (x^3)?
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  5. #5
    Member Miss's Avatar
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    Quote Originally Posted by Archduke01 View Post
    But, then what is my v if my dv is cos (x^3)?
    I told you can not find v in terms of the elementary functions.
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  6. #6
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    \frac{1}{3}\int{ucosu}du

    dv=cosudu\ \Rightarrow\ v=sinu

    \frac{1}{3}\int{ucosu}du=\frac{1}{3}\left(usinu-\int{sinu}du\right)

    =\frac{1}{3}\left(x^3sin\left(x^3\right)--cos\left(x^3\right)\right)+C

    =\frac{1}{3}\left(x^3sin\left(x^3\right)+cos\left(  x^3\right)\right)+C
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