Answer
$2\sin^{-1}\frac{x}{2}+\frac{1}{2}x\sqrt{4-x^2}+C$
Work Step by Step
$\int \sqrt{4-x^2}dx$
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 (*),
$=2\sin^{-1}\frac{x}{2}+x\sqrt{1-(\frac{x}{2})^2}+C$
$=2\sin^{-1}\frac{x}{2}+x\sqrt{1-\frac{x^2}{4}}+C$
$=2\sin^{-1}\frac{x}{2}+x\sqrt{\frac{1}{4}(4-x^2)}+C$
$=\boxed{2\sin^{-1}\frac{x}{2}+\frac{1}{2}x\sqrt{4-x^2}+C}$
