# Proving Integral of an Irrational Function

1. May 17, 2008

### PFStudent

Hey,

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

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 ${x}$ term on the outside of the radical which is preventing from evaluating this integral by normal convention.

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

In applying trigonometric substitution - consider the right triangle,

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

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?

Thanks,

-PFStudent

Last edited by a moderator: May 19, 2008
2. May 17, 2008

### 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.

3. May 17, 2008

### 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...

4. May 17, 2008

### HallsofIvy

Staff Emeritus
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 $|x|\le|a|$.

5. May 17, 2008

### csprof2000

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

6. Jun 21, 2008

### PFStudent

Hey,

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

Thanks,

-PFStudent