So what is it, in that case?

Tiiba

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!

eJavier

[tex] \int (1-x^2)^{1/2} = 1/2 ( x(1-x^2)^{1/2} + \arcsin x) [/tex]

Muzza

I'm not sure I follow, how does the integral of sqrt(1 - x^2) say anything about the transcendence of pi?

HallsofIvy

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
[itex] \int (1-x^2)^{1/2} = \int cos^2(/theta)d\theta [/itex]
But [itex]cos^2(\theta)= \frac{1}{2}(1+cos(2\theta))[/itex] so the integral becomes
[itex]\frac{1}{2}\int (1+ cos(2\theta))d\theta= \frac{1}{2}\theta+ \frac{1}{4}sin(2\theta)+ C[/itex]. 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.

Muzza

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.

Tiiba

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!

mathwonk

to integrate sqrt(1-x^2) just substitute x = sin(u), and dx = cos(u)du

Zurtex

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.