Proving Integral of an Irrational Function

[PFStudent]

Hey,

Homework Statement

(From an Integration Table)
Prove,
<br /> {\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}{|}}}<br />

Homework Equations

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

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

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

<br /> {\int}{u(x)}{d[v(x)]} = {{u(x)}{v(x)}} - {\int}{v(x)}{d[u(x)]}<br />

The Attempt at a Solution

What is bothering me about this integral is that I do not have a {x} term on the outside of the radical which is preventing from evaluating this integral by normal convention.

Let,
<br /> {I} = {\int}{\sqrt {{{a}^{2}} - {{x}^{2}}}}{dx}<br />

In applying trigonometric substitution - consider the right triangle,

So,
<br /> {{a}{\sin{\theta}}} = {\sqrt {{{a}^{2}} - {{x}^{2}}}}<br />

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

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

Thanks,

-PFStudent

 

[e(ho0n3]

I would justify it by arguing that x is the length of a side of a triangle, which is a non-negative quantity.

 

[csprof2000]

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]

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 |x|\le|a|.

 

[csprof2000]

Yeah, that's what I was originally thinking too, HallsOfIvy. But does that count as proof? I guess it should. huh.

 

[PFStudent]

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