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Question regarding an integral

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  1. Jul 4, 2016 #1
    1. The problem statement, all variables and given/known data
    I've got an equation which I need to integrate. However, integrating it and checking with the solutions, I get two different results. I get the same result as using wolfram alpha, but a different result from the book.
    If I differentiate both results, I get back to the orginal equation, so I don't know why they are different.

    2. Relevant equations
    Equation which I need to Integrate by seperation of variables: [tex]\frac{d\dot{x}^2}{dx} = \frac{2l^4 a^2 x}{(12x^2+l^2)^2}[/tex]

    Result from Book: [tex]\dot{x}^2 = \frac{ l^2 a^2 x^2}{12x^2+l^2}[/tex]
    Result from Wolfram-Alpha: [tex]\dot{x}^2 = -\frac{(a^2 l^4)}{(12 (l^2+12 x^2))}[/tex]

    3. The attempt at a solution
    If I differentiate both, I'll both get the same equation back which I integrated, I just don't understand why thats the case, since it doesn't look like they differ by a additive constant.
     
    Last edited: Jul 4, 2016
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  3. Jul 4, 2016 #2

    Ray Vickson

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    (1) Please clarify: does
    [tex] \frac{d\dot{x}^2}{dx} [/tex]
    mean
    [tex] \frac{d( d(x^2)/dt)}{dx} [/tex]
    or
    [tex] \frac{d (dx/dt)^2}{dx} [/tex]
    or
    [tex] \left( \frac{d (dx / dt) }{dx}\right)^2 \:? [/tex]
    (2) Show the steps you used in obtaining your answer.
     
    Last edited: Jul 4, 2016
  4. Jul 4, 2016 #3
    It means :[tex]\frac{d\dot{x}^2}{dx} = \frac{d}{dx}\left(\frac{dx}{dt}\right)^2[/tex]

    For the result from the book there where no steps shown.

    For my result: [tex]\frac{d\dot{x}^2}{dx} = \frac{2l^4a^2x}{(12x^2+l^2)^2} \Longrightarrow \int d\dot{x}^2 =2 l^4a^2\int \frac{x}{(12x^2+l^2)^2} dx[/tex]

    Apply Substitution: [tex]u = 12x^2 +l^2 \,\,\,\,\,\,\, \frac{du}{dx} = 24x \Rightarrow dx = \frac{1}{24x}du[/tex]

    [tex]\int d\dot{x}^2 = \dot{x}^2 = l^4a^2 \int\frac{1}{24u^2}du = -2l^4a^2\frac{1}{24} u^{-1} + c = -l^4a^2\frac{1}{12}\frac{1}{12x^2+l^2}+c[/tex]

    There were some starting conditions which would lead to c = 0
     
  5. Jul 4, 2016 #4

    Ray Vickson

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    You are correct, and the book is wrong (as can be checked by differentiating the answer---a step you should always perform).
     
  6. Jul 4, 2016 #5
    I did perform that step as I wrote in my post and I get the same equation, though my derivative might be wrong.

    Let's differentiate the solution from the book with respect to x:

    [tex]\frac{d}{dx}\left(\frac{l^2a^2x^2}{12x^2+l^2}\right) = l^2a^2 \frac{d}{dx}\left(\frac{x^2}{12x^2+l^2}\right)[/tex]

    Using the quotient rule I get: [tex]l^2a^2 \frac{2x(12x^2+l^2)-24x^3}{(12x^2+l^2)^2} = l^2a^2\frac{24x^3-2xl^2-24x^3}{(12x^2+l^2)^2} = \frac{2l^4a^2x}{(12x^2+l^2)^2}[/tex] which is the same as the orginal function which I integrated
     
  7. Jul 4, 2016 #6

    vela

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    Even though it doesn't look like it at first glance, the two results do, in fact, differ by just an additive constant:
    $$\frac{a^2l^2}{12}-\frac{a^2l^4}{12(12x^2+l^2)} = \frac{l^2a^2x^2}{12x^2+l^2}$$
     
  8. Jul 4, 2016 #7

    Ray Vickson

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    Oops, my mistake in #4.
     
  9. Jul 5, 2016 #8
    Ah I see, thanks very much both of you!
     
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