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Proving Integral of an Irrational Function

  1. May 17, 2008 #1

    1. The problem statement, all variables and given/known data
    (From an Integration Table)
    {\int}{\sqrt {{{a}^{2}} - {{x}^{2}}}}{dx} = {{\frac {1}{2}}{\left(}{{{x}{\sqrt {{{a}^{2}} - {{x}^{2}}}}} + {{{a}^{2}}{\arcsin{\frac {x}{a}}}}}{\right)}}{,}{\,}{\,}{\,}{\,}{\,}{\,}{{{|}{x}{|}}{\leq}{{|}{a}{|}}}

    2. Relevant equations
    Knowledge of Trigonometric Substitution for Integration ("Backwards" and "Forwards").

    Knowledge of integration techniques involving the form,
    {\sqrt {{{a}^{2}} - {{x}^{2}}}}

    Integration by Parts (IBP),
    {\int}{d{\left[}{u(x)}{v(x)}{\right]}} = {\int}{\biggl[}{u(x)}{d[v(x)]} + {v(x)}{d[u(x)]}{\biggl]}

    {\int}{u(x)}{d[v(x)]} = {{u(x)}{v(x)}} - {\int}{v(x)}{d[u(x)]}

    3. The attempt at a solution
    What is bothering me about this integral is that I do not have a [itex]{x}[/itex] term on the outside of the radical which is preventing from evaluating this integral by normal convention.

    {I} = {\int}{\sqrt {{{a}^{2}} - {{x}^{2}}}}{dx}

    In applying trigonometric substitution - consider the right triangle,

    {{a}{\sin{\theta}}} = {\sqrt {{{a}^{2}} - {{x}^{2}}}}

    Solving by "forward" substitution,
    {x} = {\pm}{\sqrt {{{a}^{2}}{{\cos}^{2}{\theta}}}}

    However, how would you justify which root to take: the positive one or the negative one?


    Last edited by a moderator: May 19, 2008
  2. jcsd
  3. May 17, 2008 #2
    I would justify it by arguing that x is the length of a side of a triangle, which is a non-negative quantity.
  4. May 17, 2008 #3
    I'm pretty sure all you have to do is...

    x = a sin(t)
    dx = a cos(t) dt

    I = a S sqrt(1 - sin^2 t) * a cos(t) dt = a^2 S cos^2 t dt

    Now just use a double-angle identity to evaluate the integral and then perform the appropriate substitutions to get back to the solution to the original problem. Not too bad...
  5. May 17, 2008 #4


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    Staff Emeritus
    Science Advisor

    Since you are given the integration formula and asked to prove it, you could just differentiate. And, of course, the given formula is defined only for [itex]|x|\le|a|[/itex].
  6. May 17, 2008 #5
    Yeah, that's what I was originally thinking too, HallsOfIvy. But does that count as proof? I guess it should. huh.
  7. Jun 21, 2008 #6

    Ok, that makes sense - I kind of thought that reasoning as well. Thanks.


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