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This is probably a very stupid question, but I just learned about integrals so I was wondering what happens if we calculate the integral of $\sqrt{1 - x^2}$ from $-1$ to $1$.

We would get the surface of the semi-circle, which would equal to $\pi/2$.

Would it be possible to calculate $\pi$ this way?

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Yes, that is precisely correct (although I'm not sure what you mean by "without using pi"). – Qiaochu Yuan Feb 19 '11 at 12:23
Well, I don't know how to calculate the integral of $\sqrt{1-x^2}$, but is calculating the surface possible this way using integrals, instead of using pi itself? So as a workaround, so to say. – pimvdb Feb 19 '11 at 12:24
This is not a stupid question at all. First of all, there is no way to "compute $\pi$ precisely". Figuring out approximations for $\pi$, say by estimating the area of a circle, is a famous part of mathematics. See – Sam Nead Feb 19 '11 at 19:13
up vote 5 down vote accepted

In fact, the indefinite integral of $\sqrt{1-x^2}$ is $\frac12(x\sqrt{1-x^2} + \arcsin{x}) + C$, so you are actually "using" $\pi$ in the arcsine if you solve this somehow symbolically, as

$$\int_{-1}^1 \sqrt{1-x^2}\,\mathrm dx = \arcsin 1 = \frac{\pi}{2}$$

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If you want to calculate $\pi$ in this way, note that the expansion of

$$\sqrt{1-x^2} = 1 - \sum_{n=1}^\infty \frac{(2n)!}{(2n-1)2^{2n}(n!)^2} x^{2n} $$

and so if we integrate term by term and evaluate from $-1$ to $1$ we will end up with the following formula for $\pi$:

$$ \pi = 4 \left\lbrace 1 - \sum_{n=1}^\infty \frac{(2n)!}{(4n^2-1)2^{2n}(n!)^2} \right\rbrace .$$

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You can also refer this thread:

$$ \int\limits_{0}^{1} \frac{x^{5}(1-x)^{6}(197+462x^{2})}{530(1+x^{2})} + \frac{333}{106}= \pi$$

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Yes, this integral converges to $\pi/2$. If you evaluate the integral numerically, with your favorite integration scheme, you can compute digits of $\pi$.

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Thanks. Is it perhaps possible to express $\pi/2$ using square roots or something like that, instead of $\pi$ itself? – pimvdb Feb 19 '11 at 12:27
If you mean roots of rational numbers, then no, because pi is transcendental and combinations of roots of rationals are algebraic – Chris Card Feb 19 '11 at 12:30
There is the reciprocal of Viète's formula $\frac{\pi}{2}= \frac{2}{\sqrt2}\cdot \frac{2}{\sqrt{2+\sqrt2}}\cdot \frac{2}{\sqrt{2+\sqrt{2+\sqrt2}}}\cdots$ but that is an infinite product. – Henry Feb 19 '11 at 14:03
There are many formulas shown on under the heading Computation in the computer age that involve powers and roots and converge very quickly – Ross Millikan Feb 19 '11 at 15:29

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