The area of circle

$$2\sqrt{r^2-x^2}=2r\sqrt{1-\left(\frac xr\right)^2}$$

Substitute

$$x=r\sin t\;,\;\;dx=r\cos t\,dt\implies 2r\sqrt{1-\left(\frac xr\right)^2}dx=2r^2\cos^2t\,dt\implies$$

$$\int\limits_{-r}^r2\sqrt{r^2-x^2}dx=2r^2\int\limits_{-\pi/2}^{\pi/2}\cos^2t\,dt=\left.r^2\left(t+\sin t\cos t\right)\right|_{-\pi/2}^{\pi/2}=\pi r^2$$

Hint: try a trigonometric substitution. In particular, try setting $x = r \sin \theta$.

Also, note the identity: $$ \cos^2 \theta = \frac 12 (1 + \cos(2\theta)) $$