## Calculus, 10th Edition (Anton)

$2\sin^{-1}\frac{x}{2}+\frac{1}{2}x\sqrt{4-x^2}+C$
$\int \sqrt{4-x^2}dx$ Let $x=2\sin \theta$, and $dx=2\cos \theta d\theta$. $=\int\sqrt{4-(2\sin \theta)^2}*2\cos \theta d\theta$ $=2\int\sqrt{4-4\sin^2\theta}\cos\theta d\theta$ $=2\int\sqrt{4\cos^2\theta}\cos\theta d\theta$ $=2\int 2\cos\theta\cos\theta d\theta$ $=4\int \cos^2\theta d\theta$ $=4\int\frac{1+\cos 2\theta}{2} d\theta$ $=2\int (1+\cos 2\theta) d\theta$ Let $u=2\theta$, and $du=2d\theta$. $=2\int(1+\cos u) \frac{du}{2}$ $=\int (1+\cos u)du$ $=u+\sin u+C$ $=2\theta+\sin2\theta+C$ (*) Now we want to express everything in terms of $\theta$. Since $x=2\sin \theta$, $\frac{x}{2}=\sin\theta$, so $\theta=\sin^{-1}\frac{x}{2}$. Also, $\sin 2\theta=2\sin \theta\cos\theta=2\sin\theta\sqrt{1-\sin^2\theta}$ $=2\sin\theta\sqrt{1-(\frac{2\sin\theta}{2})^2}=x\sqrt{1-(\frac{x}{2})^2}$ Going back to (*), $=2\sin^{-1}\frac{x}{2}+x\sqrt{1-(\frac{x}{2})^2}+C$ $=2\sin^{-1}\frac{x}{2}+x\sqrt{1-\frac{x^2}{4}}+C$ $=2\sin^{-1}\frac{x}{2}+x\sqrt{\frac{1}{4}(4-x^2)}+C$ $=\boxed{2\sin^{-1}\frac{x}{2}+\frac{1}{2}x\sqrt{4-x^2}+C}$