√(4-x²)의 정적분(구간 0에서 1)

$x = 2 \sin \theta$ $\displaystyle \left( - \frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} \right)$로 놓으면 $dx = 2 \cos \theta d \theta$이고 $x = 0$일 때 $\theta = 0$, $x = 1$일 때 $\displaystyle \theta = \frac{\pi}{6}$ $\displaystyle \sqrt{4 - x ^{2}} = 2 \cos \theta$ ($\cos \theta \geq 0$)이므로 $\displaystyle \int _{0} ^{1} \sqrt{4 - x ^{2}} dx = \int _{0} ^{\frac{\pi}{6}} 4 \cos ^{2} \theta d \theta = \int _{0} ^{\frac{\pi}{6}} 2 ( 1 + \cos 2 \theta ) d \theta$ $\displaystyle = \left[ 2 \theta + \sin 2 \theta \right] _{0} ^{\frac{\pi}{6}} = \frac{\pi}{3} + \sin \frac{\pi}{3} = \frac{\pi}{3} + \frac{\sqrt{3}}{2}$ 따라서 구하는 값은 $\displaystyle \frac{\pi}{3} + \frac{\sqrt{3}}{2}$ ②