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Math Help - Indefinite integrals

  1. #1
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    Exclamation Indefinite integrals

    Question:
     \int x^3.(2-x^2)^7

     u=2-x^2
     \int x^3 .u^7
     du=-2x
    I dont know what to do next.
    Please answer in steps ,Thank you
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  2. #2
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    Quote Originally Posted by mj.alawami View Post
    Question:
     \int x^3.(2-x^2)^7

     u=2-x^2
     \int x^3 .u^7
     du=-2x
    I dont know what to do next.
    Please answer in steps ,Thank you
    For this you should use integration by parts.

    Remember that \int{u\,dv} = uv - \int{v\,du}.

    In this case, let u = x^2 so that du = 2x

    Let dv = x(2 - x^2)^7 so that v = \int{x(2 - x^2)^7\,dx}.

    v = -\frac{1}{2}\int{-2x(2 - x^2)^7}

     = -\frac{1}{2}\cdot\frac{1}{8}(2 - x^2)^8

     = -\frac{1}{16}(2 - x^2)^8.



    So \int{x^3(2 - x^2)^7\,dx} = -\frac{1}{16}x^2(2 - x^2)^8 - \int{-\frac{1}{8}x(2 - x^2)^8\,dx}

     = -\frac{1}{16}x^2(2 - x^2)^8 - \frac{1}{16}\int{-2x(2 - x^2)^8\,dx}

     = -\frac{1}{16}x^2(2 - x^2)^8 - \frac{1}{16}\cdot\frac{1}{9}(2 - x^2)^9 + C

     = -\frac{1}{16}x^2(2 - x^2)^8 - \frac{1}{144}(2 - x^2)^9 + C.
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  3. #3
    MHF Contributor chisigma's Avatar
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    The substitution u=2-x^{2} is itself correct, but the complete procedure is...

    u=2-x^{2} \rightarrow x^{3} = (2-u)^{\frac{3}{2}} (1)

    dx= -\frac{du}{2x}= -\frac{du}{2\cdot (2-u)^{\frac{1}{2}}} (2)

    ... so that the integral becomes...

    - \frac{1}{2}\cdot \int u^{7} \cdot (2-u)\cdot du (3)

    ... that really seemes more easy than the original one...

    Kind regards

    \chi \sigma
    Last edited by chisigma; January 9th 2010 at 07:13 AM. Reason: Error of sign in (3)... thanks to HallsofIvy
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  4. #4
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    Quote Originally Posted by mj.alawami View Post
    Question:
     \int x^3.(2-x^2)^7

     u=2-x^2
     \int x^3 .u^7
     du=-2x
    I dont know what to do next.
    Please answer in steps ,Thank you
    Again, "du= -2x" is not correct. du= -2xdx so -\frac{1}{2}du= xdx. You need to learn to write that correctly!
    Rewrite \int x^2(2-x^2)^7 dx as \int x^2(2- x^2)^7(xdx) Now (2- x^2)^7 becomes u^7, x dx becomes -\frac{1}{2} du and x^2 becomes 2- u.
    \int x^3(2- x^2)^7 dx becomes -\frac{1}{2}\int (2-u)u^7 du= -\frac{1}{2}\int (2u^7- u^8) du
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