# Proving Integral of an Irrational Function

by PFStudent
Tags: function, integral, irrational, proving
 P: 171 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
 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.
 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...
Math
Emeritus
Thanks
PF Gold
P: 38,705

## 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 $|x|\le|a|$.
 P: 288 Yeah, that's what I was originally thinking too, HallsOfIvy. But does that count as proof? I guess it should. huh.
P: 171
Hey,

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

Thanks,

-PFStudent

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