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Math Help - Integral help

  1. #1
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    Integral help

    Is this a simple negative exponent, power chain rule

    Or an inverse sin integral?

    And if it's an inverse sin integral, how do i get rid of the first x under the radical?

    \int \frac{dx}{\sqrt{3x-x^2}}
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  2. #2
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    Quote Originally Posted by penguinpwn View Post
    Is this a simple negative exponent, power chain rule

    Or an inverse sin integral?

    And if it's an inverse sin integral, how do i get rid of the first x under the radical?

    \int \frac{dx}{\sqrt{3x-x^2}}
    it's a pain-in-the-### arcsin ...

    3x - x^2 =

    -\left(x^2 - 3x + \frac{9}{4} - \frac{9}{4}\right) =<br />

    -\left[\left(x - \frac{3}{2}\right)^2 - \left(\frac{3}{2}\right)^2\right] =

    \left(\frac{3}{2}\right)^2 - \left(x - \frac{3}{2}\right)^2

    now use the general form of the integral ...

    \int \frac{du}{\sqrt{a^2 - u^2}} = \arcsin\left(\frac{u}{a}\right) + C
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  3. #3
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    Quote Originally Posted by skeeter View Post
    it's a pain-in-the-### arcsin ...

    3x - x^2 =

    -\left(x^2 - 3x + \frac{9}{4} - \frac{9}{4}\right) =<br />

    -\left[\left(x - \frac{3}{2}\right)^2 - \left(\frac{3}{2}\right)^2\right] =

    \left(\frac{3}{2}\right)^2 - \left(x - \frac{3}{2}\right)^2

    now use the general form of the integral ...

    \int \frac{du}{\sqrt{a^2 - u^2}} = \arcsin\left(\frac{u}{a}\right) + C
    Where did the 9/4 came from?
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  4. #4
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    Quote Originally Posted by penguinpwn View Post
    Where did the 9/4 came from?
    ever heard of completing the square?
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  5. #5
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    Quote Originally Posted by skeeter View Post
    ever heard of completing the square?
    Ooooohhhhhhh, haha I see it now.

    Thanks for the help!
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