<a href="http://www.codecogs.com/eqnedit.php?latex=I=\int^{\sqrt{3}}_{-1}\frac{x^2}{\sqrt{4-x^2}}\\ $Let $x=2\sin \theta\\ dx=2\cos \theta d\theta\\ x=-1\Rightarrow \theta = -\frac{\pi}{6}\\ x=3 \Rightarrow \theta = \frac{\pi}{3}\\ ~\\ I=4\int^{\frac{\pi}{3}}_{-\frac{\pi}{6}} \frac{\sin^2\theta\cos \theta d\theta}{\cos \theta}\\ =2\int^{\frac{\pi}{3}}_{-\frac{\pi}{6}} (1-\cos 2\theta)d\theta\\ =2\left [\theta-\frac{1}{2}\sin 2\theta \right ]^{\frac{\pi}{3}}_{-\frac{\pi}{6}}\\ =2\left ( \frac{\pi}{3}-\frac{\sqrt{3}}{4}@plus;\frac{\pi}{6}-\frac{\sqrt{3}}{4}\right )\\ =\pi-\sqrt{3}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?I=\int^{\sqrt{3}}_{-1}\frac{x^2}{\sqrt{4-x^2}}\\ $Let $x=2\sin \theta\\ dx=2\cos \theta d\theta\\ x=-1\Rightarrow \theta = -\frac{\pi}{6}\\ x=3 \Rightarrow \theta = \frac{\pi}{3}\\ ~\\ I=4\int^{\frac{\pi}{3}}_{-\frac{\pi}{6}} \frac{\sin^2\theta\cos \theta d\theta}{\cos \theta}\\ =2\int^{\frac{\pi}{3}}_{-\frac{\pi}{6}} (1-\cos 2\theta)d\theta\\ =2\left [\theta-\frac{1}{2}\sin 2\theta \right ]^{\frac{\pi}{3}}_{-\frac{\pi}{6}}\\ =2\left ( \frac{\pi}{3}-\frac{\sqrt{3}}{4}+\frac{\pi}{6}-\frac{\sqrt{3}}{4}\right )\\ =\pi-\sqrt{3}" title="I=\int^{\sqrt{3}}_{-1}\frac{x^2}{\sqrt{4-x^2}}\\ $Let $x=2\sin \theta\\ dx=2\cos \theta d\theta\\ x=-1\Rightarrow \theta = -\frac{\pi}{6}\\ x=3 \Rightarrow \theta = \frac{\pi}{3}\\ ~\\ I=4\int^{\frac{\pi}{3}}_{-\frac{\pi}{6}} \frac{\sin^2\theta\cos \theta d\theta}{\cos \theta}\\ =2\int^{\frac{\pi}{3}}_{-\frac{\pi}{6}} (1-\cos 2\theta)d\theta\\ =2\left [\theta-\frac{1}{2}\sin 2\theta \right ]^{\frac{\pi}{3}}_{-\frac{\pi}{6}}\\ =2\left ( \frac{\pi}{3}-\frac{\sqrt{3}}{4}+\frac{\pi}{6}-\frac{\sqrt{3}}{4}\right )\\ =\pi-\sqrt{3}" /></a>
