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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").2011-02-19
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    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.2011-02-19
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    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 http://en.wikipedia.org/wiki/Numerical_approximations_of_%CF%802011-02-19
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    Find Pi using integrals in Geogeobra or Desmos: https://t.co/b3Cu0FVBYi and https://t.co/HbNeArfQWP2017-04-09

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There is one more that is not necessarily an integral but is interesting nonetheless $\tan^{-1} x=x-\frac{x^3}{3}+\frac{x^5}{5}...\frac{(-1)^{n}x^{2n+1}}{2n+1}$ so $\pi=4\tan^{-1} 1=4\left (1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}...\right )=$ $4\sum_{n=0}^{\infty}\frac{(-1)^{n}}{2n+1}$ but as an integral... $4\int_{0}^{1}\frac{1}{1+x^2}\text{d}x=\pi$

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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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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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    Or $$\int \limits_{-1}^{1} \frac{1}{\sqrt{1-x^2}}\,{\rm d}x = \pi $$ since the indefinite integral is $\arcsin(x)$2016-12-29
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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?2011-02-19
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    If you mean roots of rational numbers, then no, because pi is transcendental and combinations of roots of rationals are algebraic2011-02-19
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    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.2011-02-19
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    There are many formulas shown on http://en.wikipedia.org/wiki/Pi under the heading Computation in the computer age that involve powers and roots and converge very quickly2011-02-19
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You can also refer this thread:

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