## So what is it, in that case?

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?

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## So what is it, in that case?

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?
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!
 Recognitions: Homework Help Science Advisor to integrate sqrt(1-x^2) just substitute x = sin(u), and dx = cos(u)du

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