Calculating value of $\pi$ independently using integrals.

Its famous integral :) I too researched on this some time before.

I found this, quite useful : Integral approximations to π with nonnegative integrands.

From wolfram I get this:

Backhouse (1995) used the identity $$\begin{aligned}I_{m, n} &= \int_{0}^{1}\frac{x^{m}(1 - x)^{n}}{1 + x^{2}}\,dx\\ &= 2^{-(m + n + 1)}\sqrt{\pi}\Gamma(m + 1)\Gamma(n + 1)\times{}_{3}F_{2}\left(1, \frac{m + 1}{2}, \frac{m + 2}{2};\frac{m + n + 2}{2}, \frac{m + n + 3}{2}; -1\right)\\ &= a + b\pi + c\log 2\end{aligned}$$ for positive integers $m$ and $n$ and where $a, b$ and $c$ are rational constants to generate a number of formulas for $\pi$. In particular, if $$2m - n \equiv 0\pmod{4}$$ then $c = 0$ (Lucas 2005).

This integral has a series equivalent

$$\frac{22}{7}-\pi=\sum_{k=1}^\infty \frac{240}{(4k+1)(4k+2)(4k+3)(4k+5)(4k+6)(4k+7)}$$

More generally, $$\sum_{k=m}^\infty \frac{240}{(4k+1)(4k+2)(4k+3)(4k+5)(4k+6)(4k+7)}=\int_0^1 \frac{x^{4m}(1-x)^4}{1+x^2}dx$$

which gives closer rational approximations to $\pi$ by adding more terms of the summation.

https://math.stackexchange.com/a/1657416/134791

A similar result is obtained changing the denominator to $\frac{1-x^8}{1-x^2}=1+x^2+x^4+x^6$.

$$ \frac{2\sqrt{2}}{3} \int_0^1 \frac{x^4 (1-x)^4} {1+x^2+x^4+x^6} dx = \frac{20 \sqrt{2}}{9}-\pi$$

Other interesting denominators are $1+x+x^2$, $\frac{1-x^6}{1-x^2}$ and $\frac{1-x^{12}}{1-x^2}$, which lead to approximations involving $\sqrt{3}$ and different exponents in the numerator.