This is a typical problem which can be solved using trigonometric substitution. Let $x=\sin\theta$, then $\theta\in[0, \pi/2]$.

$$\begin{align*} &\int_0^1 \sqrt{1-x^2}d{x}\\ &=\int_{0}{\frac{\pi}{2}}\cos\theta d{\sin\theta}\\&=\int_{0}^{1}\cos^2\theta d{\theta}\\&=\int_{0}^{1}\frac{1+\cos 2\theta}{2}d{\theta}\\&=\frac{1}{2}\left(\left.\theta\right|_{0}^{\frac{\pi}{2}}+\int_{0}^{\frac{\pi}{2}}\cos 2\theta d{\theta}\right)\\&=\frac{1}{2}\left(\frac{\pi}{2}+\frac{1}{2}\left.\sin 2\theta\right|_0^{\frac{\pi}{2}}\right)\\&=\boxed{\frac{\pi}{4}}\end{align*}$$

This result can also be obtained by noting that this expression equals the area of a quarter unit circle.