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Math Help - Simple Fraction Integration

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    Simple Fraction Integration

    Q: \displaystyle\int \frac{x}{\sqrt{4-x^2}} \, \mathrm{d}x

    Does that just equal \sqrt{4-x^2}? If so, why?
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    Quote Originally Posted by Air View Post
    Q: \displaystyle\int \frac{x}{\sqrt{4-x^2}} \, \mathrm{d}x

    Does that just equal \sqrt{4-x^2}? If so, why?

    Let u = 4 - x^2

    Then \frac{du}{dx} = -2x


    \int \frac{x}{\sqrt{4-x^2}} dx

    = -\frac{1}{2} \int \frac{1}{\sqrt{u}} du

    = -\frac{1}{2} \int (u)^{- \frac{1}{2}} du

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

    = -(\sqrt{4 - x^2})
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  3. #3
    Moo
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    Hello,

    Quote Originally Posted by Air View Post
    Q: \displaystyle\int \frac{x}{\sqrt{4-x^2}} \, \mathrm{d}x

    Does that just equal \sqrt{4-x^2}? If so, why?
    \frac{x}{\sqrt{4-x^2}}=\frac{2x}{2 \sqrt{4-x^2}}

    Here, it's like \frac{-u'(x)}{2 \sqrt{u(x)}}

    An antiderivative of this is -\sqrt{u(x)}=-\sqrt{4-x^2}
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    When I use substitution of u^2 = 4 - x^2, I seem to get an integral of \int - \frac{1}{u} \ \mathrm{d}u, but that would be \ln u. Doesn't u^2 = 4 - x^2 substitution work?
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    Moo
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    Quote Originally Posted by Air View Post
    When I use substitution of u^2 = 4 - x^2, I seem to get an integral of \int - \frac{1}{u} \ \mathrm{d}u, but that would be \ln u. Doesn't u^2 = 4 - x^2 substitution work?
    If you do this substitution :

    2 udu=-2x dx -> dx=-\frac ux

    -> \int \frac{x}{\sqrt{4-x^2}} dx=\int \frac{u}{u} du=u=\pm \sqrt{4-x^2}

    It's quite dangerous because of this \pm
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  6. #6
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    Quote Originally Posted by Moo View Post
    If you do this substitution :

    2 udu=-2x dx -> dx=-\frac ux

    -> \int \frac{x}{\sqrt{4-x^2}} dx=\int \frac{u}{u} du=u=\pm \sqrt{4-x^2}

    It's quite dangerous because of this \pm
    Why did you get \int \frac{u}{u} \ \mathrm{d}u? Where did the u in the numerator come from?
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  7. #7
    Moo
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    Quote Originally Posted by Air View Post
    Why did you get \int \frac{u}{u} \ \mathrm{d}u? Where did the u in the numerator come from?
    From dx

    \int \frac{x}{\sqrt{4-x^2}} \underbrace{dx}_{\frac{-u}{x} \cdot du}=\int \frac{x}{u} \cdot \frac{-u}{x} du





    Typo
    Quote Originally Posted by Moo View Post
    If you do this substitution :

    2 udu=-2x dx -> dx=-\frac ux

    -> \int \frac{x}{\sqrt{4-x^2}} dx={\color{red}-}\int \frac{u}{u} du={\color{red}-}u=\pm \sqrt{4-x^2}

    It's quite dangerous because of this \pm
    Last edited by Moo; May 8th 2008 at 08:46 AM.
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