What is the Antiderivative of Sqrt(1-x^2)?

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Discussion Overview

The discussion revolves around finding the antiderivative of the function sqrt(1-x^2), exploring its implications, and the relationship between this integral and the nature of the number pi. The scope includes mathematical reasoning and integration techniques.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant expresses frustration in finding the antiderivative of sqrt(1-x^2) and attempts to connect it to the algebraic nature of pi.
  • Another participant provides a proposed antiderivative, stating it as 1/2 ( x(1-x^2)^{1/2} + arcsin x).
  • A participant questions the connection between the integral of sqrt(1-x^2) and the transcendence of pi, seeking clarification.
  • Another participant explains a substitution method using x = sin(θ) to integrate sqrt(1-x^2), leading to an expression involving π, but emphasizes that this does not imply anything about pi's algebraic status.
  • One participant speculates about the possibility of expressing pi as a difference of algebraic formulas, suggesting that if it could be done, pi would be algebraic, but notes the presence of inverse trigonometric functions complicates this.
  • Another participant reflects on the amusing observation that the area under a reciprocal of a quadratic equation is a multiple of pi.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the implications of the integral regarding the nature of pi, with some expressing confusion and others clarifying that the integral does not provide evidence for pi being algebraic or transcendental.

Contextual Notes

There are unresolved assumptions regarding the nature of pi and its relationship to algebraic expressions, as well as the mathematical steps involved in the integration process.

Tiiba
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I tried to find an antiderivative of sqrt(1-x^2), integrate a unit semicircle (which should be pi/2), and thus prove that pi is algebraic. Since pi is not algebraic, I failed miserably.

But that equation is certainly integrable from -1 to 1. So what is its antiderivative? I tried to find it on the web, but I couldn't.

*Hits the sack*

Take this, you stupid sack!
 
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\int (1-x^2)^{1/2} = 1/2 ( x(1-x^2)^{1/2} + \arcsin x)
 
I tried to find an antiderivative of sqrt(1-x^2), integrate a unit semicircle (which should be pi/2), and thus prove that pi is algebraic. Since pi is not algebraic, I failed miserably.

I'm not sure I follow, how does the integral of sqrt(1 - x^2) say anything about the transcendence of pi?
 
It doesn't. Why did you think it does?

To integrate (1-x2)1/2, let x= sin(θ) so that dx= cos(θ) and (1-x2)1/2 becomes cos(θ). Then
\int (1-x^2)^{1/2} = \int cos^2(/theta)d\theta
But cos^2(\theta)= \frac{1}{2}(1+cos(2\theta)) so the integral becomes
\frac{1}{2}\int (1+ cos(2\theta))d\theta= \frac{1}{2}\theta+ \frac{1}{4}sin(2\theta)+ C. If you evaluate that between 0 and π, the "sin" part is 0 at both ends so you get (1/2)π

That still tells you nothing about whether pi is algebraic or transcendental.
 
It doesn't. Why did you think it does?

:confused: I never claimed or believed the integral said anything about the transcendence of pi, I was looking for the reason why Tiiba thought it did.
 
If I could find an antiderivative of pi that would have only normal operations in it, roots, exponents, addition, all the nice things I learned in algebra, and pi was a difference of two such formulas, then pi, being just twice that number, would be agebraic, right?

But it just had to contain some arcsin thing in it.

Repent, arcsinner!
 
Last edited:
to integrate sqrt(1-x^2) just substitute x = sin(u), and dx = cos(u)du
 
Tiiba said:
If I could find an antiderivative of pi that would have only normal operations in it, roots, exponents, addition, all the nice things I learned in algebra, and pi was a difference of two such formulas, then pi, being just twice that number, would be agebraic, right?

But it just had to contain some arcsin thing in it.

Repent, arcsinner!
Unfortunately you'll just end up getting inverse trigonometric functions when you try things like that and Pi is involved in the answer.

During one of my exams quite recently it amused me how the area under the reciprocal of a quadratic equation was a multiple of Pi.
 

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