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Proving Integral of an Irrational Function

  • Thread starter PFStudent
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  • #1
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Hey,

Homework Statement


(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]

Homework 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]

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,
327047_Trigonometry.JPG


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:

Answers and Replies

  • #2
1,357
0
I would justify it by arguing that x is the length of a side of a triangle, which is a non-negative quantity.
 
  • #3
287
0
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...
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,793
922
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].
 
  • #5
287
0
Yeah, that's what I was originally thinking too, HallsOfIvy. But does that count as proof? I guess it should. huh.
 
  • #6
170
0
Hey,

I would justify it by arguing that x is the length of a side of a triangle, which is a non-negative quantity.
Ok, that makes sense - I kind of thought that reasoning as well. Thanks.

Thanks,

-PFStudent
 

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