More than one way

$\begin{aligned} I & = \int_0^1 \sqrt{1-x^2} dx & \small \color{#3D99F6} \text{Let } x =\sin \theta \implies dx = \cos \theta \ d\theta \\ & = \int_0^\frac \pi 2 \cos^2 \theta \ d\theta \\ & = \int_0^\frac \pi 2 \frac 12 \left(1+\cos (2 \theta)\right) \ d\theta \\ & = \frac x2 + \frac {\sin 2\theta}4 \ \bigg|_0^\frac \pi 2 \\ & = \boxed{\frac \pi 4} \end{aligned}$

There is more than one way to solve this problem! I am going to show one of them!

According to the definition of the integral, integral is the area under curve from a to b. Let y,

y = $\displaystyle\int_0^1 \sqrt{1-x^2}$ , y > 0

Since both sides are non negative, we can square both sides.

$y^2=1-x^2$ Rearrange the problem

$y^2+x^2=1$

Famous circle equations! with the center at the origin and radius 1. Area of the circle = $\pi r^2$ .

As we integrate only from 0 to 1, and we disregard area where y<0, we only need 1/4 of the circle. $\boxed{\frac{\pi}{4}}$