More to it than it seems

$\begin{aligned} I & = 2 \int_{-1}^1 \sqrt{1-x^2} dx & \small \blue{\text{Let }x = \sin \theta \implies dx = \cos \theta \ d \theta} \\ & = 2 \int_{-\frac \pi 2}^\frac \pi 2 \cos^2 \theta \ d\theta \\ & = \int_{-\frac \pi 2}^\frac \pi 2 (1 + \cos 2 \theta) \ d\theta \\ & = \left[ \theta + \frac {\sin 2 \theta}2 \right]_{-\frac \pi 2}^\frac \pi 2 \\ & = \pi \approx \boxed{3.14} \end{aligned}$

If you observe the plot of this function, you can see that it is a semicircle with a radius of 1:

We can use this to make the calculation easier because we can now use $\frac{1}{2}\pi r^2$ instead of finding the antiderivative.

$\Huge 2 \left( \frac{1}{2}\pi \times 1^2 \right) = \boxed{\pi}$