Đặt: $x=\sqrt2\sin t,t\in\left[-\dfrac{\pi}{2};\dfrac{\pi}{2}\right] \Rightarrow dx=\sqrt2\cos tdt$
Ta có:
$\int\limits_{-1}^1\sqrt{2-x^2}dx$
$=\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sqrt{2-2\sin^2t}.\sqrt2\cos tdt$
$=\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}}2\cos^2tdt$
$=\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}}(\cos2t+1)dt$
$=\left(\dfrac{\sin2t}{2}+t\right)\left|\begin{array}{l}\dfrac{\pi}{4}\\-\dfrac{\pi}{4}\end{array}\right.=1+\dfrac{\pi}{2}$