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Proving Integral of an Irrational Function

  1. May 17, 2008 #1
    Hey,

    1. The problem statement, all variables and given/known data
    (From an Integration Table)
    Prove,
    [tex]
    {\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}{|}}}
    [/tex]

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

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

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

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

    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.

    Let,
    [tex]
    {I} = {\int}{\sqrt {{{a}^{2}} - {{x}^{2}}}}{dx}
    [/tex]

    In applying trigonometric substitution - consider the right triangle,
    [​IMG]

    So,
    [tex]
    {{a}{\sin{\theta}}} = {\sqrt {{{a}^{2}} - {{x}^{2}}}}
    [/tex]

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

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

    Thanks,

    -PFStudent
     
    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

    HallsofIvy

    User Avatar
    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
    Hey,

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

    Thanks,

    -PFStudent
     
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