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Thread: Integral

  1. #1
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    Integral

    Find $\displaystyle \displaystyle \int\frac{a(a^2-x^2)}{\sqrt{(a^2+ax+x^2)(a^2-ax+x^2)^3}}\;{dx}$ (where $\displaystyle a$ is a constant, of course).
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  2. #2
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    Let me try


    Rewirte the integrand :

    $\displaystyle = \frac{a(\frac{a^2}{x^2}-1)}{\sqrt{( \frac{a^2}{x} + a + x) ( \frac{a^2}{x} - a + x )^3 } } $

    Then substitute $\displaystyle \frac{a^2}{x} + x = ay $

    $\displaystyle (\frac{a^2}{x^2} - 1 ) dx = -ady$

    The integral becomes :


    $\displaystyle - \int \frac{a^2 dy}{ \sqrt{ (ay+a)( ay-a)^3 } } $

    $\displaystyle = -\int \frac{dy}{\sqrt{(y+1)(y-1)^3 } } $

    Sub. $\displaystyle y = \frac{t+1}{t-1} $

    $\displaystyle dy = -\frac{2dt}{(t-1)^2 } $

    The integral

    $\displaystyle = \int \frac{1}{\sqrt{\frac{2t}{t-1}( \frac{2}{t-1} )^3} }~ \frac{2dt}{(t-1)^2 } $

    $\displaystyle = \frac{1}{2} \int \frac{dt}{\sqrt{t} } $

    $\displaystyle = \sqrt{t} + C $

    But $\displaystyle t = \frac{y+1}{y-1} $ we have

    $\displaystyle = \sqrt{ \frac{y+1}{y-1} } + C $

    $\displaystyle = \sqrt{ \frac{ a^2 + ax+ x^2 }{ a^2 - ax + x^2 }} + C $
    Last edited by simplependulum; Sep 5th 2010 at 01:49 AM.
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  3. #3
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    You have totally annihilated it!

    Now that I've the answer, I see that it could have been done it in the following way:

    Rewrite the integrand:

    $\displaystyle \displaystyle \frac{a(a^2-x^2)}{\sqrt{(a^2+ax+x^2)(a^2-ax+x^2)^3}}$

    $\displaystyle = \displaystyle \frac{a(a^2-x^2)}{\sqrt{(a^2+ax+x^2)}\sqrt{(a^2-ax+x^2)^3}}$

    [Math] \displaystyle = \frac{a(a^2-x^2)}{\sqrt{(a^2+ax+x^2)}\sqrt{(a^2-ax+x^2)}(a^2-ax+x^2)} [/tex]

    $\displaystyle \displaystyle = \frac{a(a^2-x^2)\sqrt{(a^2+ax+x^2)}}{(a^2+ax+x^2)((a^2-ax+x^2)\sqrt{(a^2-ax+x^2)}}$

    $\displaystyle \displaystyle = \frac{a(a^2-x^2)}{(a^2+ax+x^2)((a^2-ax+x^2)}\cdot \frac{\sqrt{(a^2+ax+x^2)}}{\sqrt{(a^2-ax+x^2)}}$

    Let $\displaystyle \displaystyle t = \frac{\sqrt{(a^2+ax+x^2)}}{\sqrt{(a^2-ax+x^2)}}$ (One would have to employ clairvoyance to spot that substitution without knowing the answer beforehand )

    Then $\displaystyle \displaystyle \ln{t} = \ln\left\{\frac{\sqrt{a^2+ax+x^2}}{\sqrt{a^2-ax+x^2}}\right\}$

    $\displaystyle \displaystyle = \ln\left(\sqrt{a^2+ax+x^2}\right)+\ln\left(\sqrt{a ^2-ax+x^2}\right)$

    $\displaystyle \displaystyle= \frac{1}{2}\ln\left(a^2+ax+x^2\right)+\frac{1}{2}\ ln\left(a^2-ax+x^2\right)$

    $\displaystyle \displaystyle \Rightarrow \left(\frac{1}{t}\right)\frac{dt}{dx} = \frac{2x+a}{2(a^2+ax+x^2)}-\frac{2x-a}{2(a^2-ax+x^2)}$

    $\displaystyle \displaystyle \Rightarrow \left(\frac{1}{t}\right)\frac{dt}{dx} = \frac{(2x+a)(a^2-ax+x^2)-(2x-a)(a^2+ax+x^2)}{2(a^2+ax+x^2)(a^2-ax+x^2)}$

    $\displaystyle \displaystyle \Rightarrow \left(\frac{1}{t}\right)\frac{dt}{dx} = \frac{2 a^3-2ax^2}{2(a^2+ax+x^2)(a^2-ax+x^2)} = \frac{ a(a^2-x^2)}{(a^2+ax+x^2)(a^2-ax+x^2)}$

    $\displaystyle \displaystyle \Rightarrow \frac{dt}{dx} = \frac{ a(a^2-x^2)}{(a^2+ax+x^2)(a^2-ax+x^2)}\cdot \frac{\sqrt{(a^2+ax+x^2)}}{\sqrt{(a^2-ax+x^2)}}$

    $\displaystyle \displaystyle \Rightarrow {dx} = \frac{(a^2+ax+x^2)(a^2-ax+x^2)}{a(a^2-x^2)}\cdot \frac{\sqrt{(a^2-ax+x^2)}}{\sqrt{(a^2+ax+x^2)}}\;{dt}$

    $\displaystyle \displaystyle \therefore I = \int \frac{a(a^2-x^2)}{\sqrt{(a^2+ax+x^2)(a^2-ax+x^2)^3}}\;{dx} $

    $\displaystyle \displaystyle = \int \bigg\{\frac{a(a^2-x^2)}{(a^2+ax+x^2)((a^2-ax+x^2)}\cdot \frac{\sqrt{(a^2+ax+x^2)}}{\sqrt{(a^2-ax+x^2)}}\bigg\}\bigg\{\frac{(a^2+ax+x^2)(a^2-ax+x^2)}{a(a^2-x^2)}\cdot \frac{\sqrt{(a^2-ax+x^2)}}{\sqrt{(a^2+ax+x^2)}}\right\}\bigg\}\;{dt }$

    The integrand cancels to 1 (since the two products are inverse to each other), therefore $\displaystyle I = t+k$.

    Hence $\displaystyle \displaystyle \int \frac{a(a^2-x^2)}{\sqrt{(a^2+ax+x^2)(a^2-ax+x^2)^3}}\;{dx} = \frac{\sqrt{(a^2+ax+x^2)}}{\sqrt{(a^2-ax+x^2)}}+k.$
    Last edited by TheCoffeeMachine; Sep 6th 2010 at 12:13 AM.
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