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Math Help - having trouble with this integral

  1. #1
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    having trouble with this integral

    can someone help using simple techniques of antidifferentiation? / substitution?
    i cant get this one
    \int \frac{x^3}{\sqrt{1-2x^2}}dx
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  2. #2
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    I have one rough solution...

    The integral to be solved is:

    \int \frac{x^3}{\sqrt{1-2x^2}} dx

    Split it up for convenience:

    \int \frac{x^3}{\sqrt{1-2x^2}} dx = \int \left( \frac{x}{\sqrt{1 - 2x^2}}\right) \left( x^2\right) dx =  \int \left( \frac{-2x}{\sqrt{1 - 2x^2}}\right) \left( -\frac{x^2}{2}\right) dx

    ...and the first bracket is now the derivative of \sqrt{1 - 2x^2}. Hence, the integrand lends itself to integration by parts.

    \int \left( \frac{-2x}{\sqrt{1 - 2x^2}}\right) \left( -\frac{x^2}{2}\right) dx

    = \left[ \sqrt{1 - 2x^2}\cdot \frac{-x^2}{2}\right] - \int \left(\sqrt{1 - 2x^2}\right) \cdot (-x) dx

    = \left[ \sqrt{1 - 2x^2}\cdot \frac{-x^2}{2}\right] - \frac{1}{4}\int \left(\sqrt{1 - 2x^2}\right) \cdot (-4x) dx

    = \left[ \sqrt{1 - 2x^2}\cdot \frac{-x^2}{2}\right] - \frac{1}{4} \cdot \frac{2}{3} \left[ (1 - 2x^2)^{3/2}\right] + C

    = \left[ \sqrt{1 - 2x^2}\cdot \frac{-x^2}{2}\right] - \frac{1}{6}  \left[ \sqrt{1 - 2x^2} (1 - 2x^2)\right] + C

    =  \sqrt{1 - 2x^2}\cdot \left[ - \frac{x^2}{2} - \frac{1}{6} + \frac{x^2}{3} \right] + C

    =  - \frac{1}{6} \sqrt{1 - 2x^2} (x^2 + 1) + C

    This agrees with The Integrator (i.e. Mathematica).
    Last edited by Pterid; June 18th 2007 at 06:53 AM.
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  3. #3
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    u = 1 - 2x^2 \quad  \Rightarrow \quad du =  - 4xdx\quad  \Rightarrow \quad xdx =  - u/4

    \int {\frac{{x^3 dx}}{{\sqrt {1 - 2x^2 } }} = \int {\frac{{\left( {\frac{{1 - u}}{2}} \right)^2 \left( {\frac{{ - du}}{4}} \right)}}{{\sqrt u }}} }
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  4. #4
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    Set u=\sqrt{1-2x^2}

    The integral becomes to

    \begin{aligned}<br />
\int {\frac{{x^3 }}<br />
{{\sqrt {1 - 2x^2 } }}~dx} ~&=~ - \frac{1}<br />
{4}\int {(1 - u^2 )~du}\\<br />
~&=~ - \frac{1}<br />
{4}\left( {u - \frac{1}<br />
{3}u^3 } \right) + k\\<br />
~&=~ - \frac{1}<br />
{6}(x^2 + 1)\sqrt {1 - 2x^2 } + k,~~k\in\mathbb{R}<br />
\end{aligned}
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  5. #5
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    ty!!!
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  6. #6
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    Quote Originally Posted by Krizalid View Post
    Set u=\sqrt{1-2x^2}

    The integral becomes to

    \begin{aligned}<br />
\int {\frac{{x^3 }}<br />
{{\sqrt {1 - 2x^2 } }}~dx} ~&=~ - \frac{1}<br />
{4}\int {(1 - u^2 )~du}\\<br />
~&=~ - \frac{1}<br />
{4}\left( {u - \frac{1}<br />
{3}u^3 } \right) + k\\<br />
~&=~ - \frac{1}<br />
{6}(x^2 + 1)\sqrt {1 - 2x^2 } + k,~~k\in\mathbb{R}<br />
\end{aligned}
    hi
    was wondering how did you substitute to (1-u^2)?
    and this is not part of integration by parts
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  7. #7
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    If u=\sqrt{1-2x^2}\implies{du=-\frac{2x}{\sqrt{1-2x^2}}}~dx

    Plus

    u=\sqrt{1-2x^2}\iff{x^2=\frac{1-u^2}2}
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  8. #8
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    hi
    im not quite sure how to do it using sub but ill using the sub of krizalid:

     \int \frac{x^3}{\sqrt{1 - 2x^2 }} dx

     \int \frac{ x . x^2}{u}\frac{\sqrt{1-2x^2}}{2x}du

    cancel x
    -\frac{1}{2} \int {\frac{\frac{1-u^2}{2}}{u}}du

    -\frac{1}{2} \int \frac{1-u^2}{2u}du
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