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Math Help - Help with integrals - chain rule

  1. #1
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    Help with integrals - chain rule

    Integrate:

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  2. #2
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    Answer: use subtitude x^2=t

    x^2 = t

    int[x^3(2-x^2)^12]dx = 1/2 int[t(2-t)^12]dt

    from here you need to use the chain rule

    =1/2 ((t^2/2)(2-t)^12 - int[(2-t)^12]dt)

    and from there just do another simple integral and replace t with x^2...
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  3. #3
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    Sorry my bad

    it should be like that after integrating in parts :

    1/2( (t(2-t)^13)/13 - int[(2-t)^12]dt)
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  4. #4
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    Quote Originally Posted by Tankado View Post
    x^2 = t

    int[x^3(2-x^2)^12]dx = 1/2 int[t(2-t)^12]dt

    from here you need to use the chain rule

    =1/2 ((t^2/2)(2-t)^12 - int[(2-t)^12]dt)

    and from there just do another simple integral and replace t with x^2...
    but theres an x^3?... how did t = x^3?
    and the method is diff from the method of my book
    and thenks for replying tankado
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  5. #5
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    Rewite as:
    (x^2(2-x^2))(xdx)

    u=2-x^2, x^2=2-u, du=-2xdx.

    (-1/2)(1-u)(u)^12 du
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  6. #6
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    Hello, ^_^Engineer_Adam^_^!

    Integrate: .x(2 - x)^{12} dx

    Integrate by parts . . .

    . . u .= .x . . . . . dv .= .x(2 - x)
    ^{12} dx

    . du .= .2x dx . . . .v .= .(-1/26)(2 - x)
    ^{13}


    And we have: . (-1/26)x(2 - x)
    ^{13} + (1/13) x(2 - x)^{13} dx

    . . Finally let: u .= .2 - x . . . etc.

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  7. #7
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    Explanation:

    Well you have been answered but i will show you my way in detailed :

    int[x^3(2-x^2)^12]dx

    t=x^2
    dt = 2xdx ==> xdx = dt/2

    int[x^3(2-x^2)^12]dx = int[x^2(2-x^2)^12]xdx = (xdx = dt/2 x^2=t) =

    = int[t(2-t)^12]1/2dt = 1/2int[t(2-t)^12]dt

    now integration by parts : int[f(t)g`(t)]dt = f(t)g(t) - int[f`(t)g(t)]dt
    let f(t) = t , g(t) = ((2-t)^13)/13
    f`(t) = 1 , g`(t) = (2-t)^12
    then :
    int[t(2-t)^12]dt = t((2-t)^13)/13 - 1/13int[(2-t)^13]dt

    from here you can go by yourself...
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  8. #8
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    Not necessary to use integration by parts, just set u = 2 - x, and it becomes very easy.
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  9. #9
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    So there is a way more easier?...No wonder i didnt understand the previuos posts...thats becuase im still taking up basic in integrals...( havent read about int by parts yet)
    could you please show to me the solution
    When u = 2-x^2
    then what is x^3?
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  10. #10
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    What I'm trying to say is


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