Can integration get the real value of $\pi$? If you take the equation of a cicle $x^2+y^2=1$ and re write it as $y=\sqrt{1-x^2}$, why can't you use $\int_{-1}^{1}(\sqrt{1-x^2})dx$ to get the real value of $\pi$ even if you end up with more transcendental numbers?
 A: I believe there is some confusion in the question. We can't calculate $\pi$ to infinite precision, because it's irrational. Therefore, anytime you see a list of digits of $\pi$, some kind of approximation has been used because if the list of digits ends, then it's not really $\pi$ but rather very close to it.
However, there are lots and lots of formulas that say $\pi$ is exactly equal to something else. Yours is one of them: $\pi = 2 \int_{-1}^1 \sqrt{1-x^2}\ dx$. If you want to use this formula to calculate $\pi$, you are going to have to make some kind of approximation: either a Riemann sum, or a series expansion, or something. That's because you can never get $\pi$ with infinite precision, since, as I said before, it's irrational and so its decimal expansion is infinite but not periodic.
It's important to note that of the many many formulas for $\pi$, some are better suited for approximations than others. A pretty popular one is Machin's formula:
$$\frac{\pi}{4} = 4 \arctan\frac15 - \arctan \frac{1}{239}$$
The reason this formula is good is that the series expansion for $\arctan x$ converges pretty fast for small $x$. In fact, using the crudest approximation $\arctan x \approx x$, we get $\pi \approx 3.18$, not bad for such a simple calculation.
