Calculate $\pi$ precisely using integrals?

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?

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 $\pi$:

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

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

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

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

You can also refer this thread:

• Is there an integral that proves $\pi > 333/106$?

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

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... $4\int_{0}^{1}\frac{1}{1+x^2}\text{d}x=\pi$