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Math Help - integrate: x/(sqrrt(1-9x^2)

  1. #1
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    integrate: x/(sqrrt(1-9x^2)

    How can I integrate x/(sqrrt(1-9x^2)?

    Thanks
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by dangerous_dave View Post
    How can I integrate x/(sqrrt(1-9x^2)?

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    You can do this two ways the first is this...recognize that

    \int\frac{x}{\sqrt{1-9x^2}}dx=\frac{-1}{9}\int\bigg(\sqrt{1-9x^2}\bigg)'dx where the answer becoems apparent


    The other way is u-sub

    Let u=1-9x^2
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  3. #3
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    {\color{blue}u = 1 -9x^{2}}
    du = -18x dx \Rightarrow {\color{red}x dx = -\frac{du}{18}}

    Looking at your integral:
    \int \frac{{\color{red}xdx}}{\sqrt{{\color{blue}1-9x^{2}}}}
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by o_O View Post
    {\color{blue}u = 1 -9x^{2}}
    du = -18x dx \Rightarrow {\color{red}x dx = -\frac{du}{18}}

    Looking at your integral:
    \int \frac{{\color{red}xdx}}{\sqrt{{\color{blue}1-9x^{2}}}}
    Ahh dear o_O whats the score again?
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  5. #5
    o_O
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    Haha uh ... let's call it a tie
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by o_O View Post
    Haha uh ... let's call it a tie
    Sounds great to me...you are my team leader after all ...::quivers in fear::
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  7. #7
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by dangerous_dave View Post
    How can I integrate x/(sqrrt(1-9x^2)?

    Thanks
    To integrate \int \frac{x}{\sqrt{1-9x^2}}\,dx, you would need to make a u-substitution: u=1-9x^2. As a result, du=-18x\,dx. Let's substitute all this stuff into the integral now:

    \int \frac{x}{\sqrt{1-9x^2}}\,dx now becomes -\frac{1}{18}\int\frac{\,du}{\sqrt{u}}=-\frac{1}{18}\int u^{-\frac{1}{2}}\,du.

    Integrate:

    -\frac{1}{18}\int u^{-\frac{1}{2}}\,du=-\frac{1}{9}\sqrt{u}+C.

    Substitute u back in and we have:


    \int \frac{x}{\sqrt{1-9x^2}}\,dx=-\frac{1}{9}\sqrt{1-9x^2}+C

    Hope that helped you out!!!
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  8. #8
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    That does help heaps, thanks
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