Proving Integral of an Irrational Function


by PFStudent
Tags: function, integral, irrational, proving
PFStudent
PFStudent is offline
#1
May17-08, 09:35 AM
P: 171
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,


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
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e(ho0n3
e(ho0n3 is offline
#2
May17-08, 10:18 AM
P: 1,370
I would justify it by arguing that x is the length of a side of a triangle, which is a non-negative quantity.
csprof2000
csprof2000 is offline
#3
May17-08, 10:21 AM
P: 288
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...

HallsofIvy
HallsofIvy is offline
#4
May17-08, 01:51 PM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,882

Proving Integral of an Irrational Function


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].
csprof2000
csprof2000 is offline
#5
May17-08, 02:52 PM
P: 288
Yeah, that's what I was originally thinking too, HallsOfIvy. But does that count as proof? I guess it should. huh.
PFStudent
PFStudent is offline
#6
Jun21-08, 08:28 PM
P: 171
Hey,

Quote Quote by e(ho0n3 View Post
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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