Proving Integral of an Irrational Function

In summary, we have a conversation about a specific integral that is proving a given formula using trigonometric substitution and integration techniques. The main concern is justifying which root to take in the substitution, with the explanation being that x is the length of a side of a triangle and is therefore non-negative. The conversation also mentions differentiating as a possible method of proof.
  • #1
PFStudent
170
0
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
 
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  • #2
I would justify it by arguing that x is the length of a side of a triangle, which is a non-negative quantity.
 
  • #3
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
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
Yeah, that's what I was originally thinking too, HallsOfIvy. But does that count as proof? I guess it should. huh.
 
  • #6
Hey,

e(ho0n3 said:
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
 

1. What is an irrational function?

An irrational function is a mathematical function that contains irrational numbers, meaning numbers that cannot be expressed as a ratio of two integers. Examples of irrational numbers include π and √2.

2. How do you prove the integral of an irrational function?

To prove the integral of an irrational function, you can use techniques such as substitution, integration by parts, or partial fractions. It is also important to understand the properties of irrational functions and how to simplify them before attempting to integrate.

3. Can you provide an example of proving the integral of an irrational function?

Sure, let's take the function ƒ(x) = √(x+1). To find the integral of this function, we can use substitution by letting u = x+1, du = dx. This gives us the integral ∫ √u du, which can be solved using the power rule. The final answer is (2/3)u^(3/2) + C, or (2/3)(x+1)^(3/2) + C.

4. What are some common mistakes when proving the integral of an irrational function?

Some common mistakes include forgetting to use the correct substitution, not simplifying the function before integrating, and making errors with algebraic manipulations. It is important to be careful and check your work when dealing with irrational functions.

5. How can proving the integral of an irrational function be useful?

Proving the integral of an irrational function can be useful in many real-world applications, such as calculating the area under a curve or determining the displacement of an object. It is also an important concept in higher level math courses, such as calculus and differential equations.

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