Let $x=2\sin\theta$, then $\mathrm{d}x = 2\cos\theta\mathrm{d}\theta$. Then
\begin{eqnarray*}
\int \sqrt{4-x^2}\mathrm{d}x &=& \int \sqrt{4-4\sin^2\theta}\cdot 2\cos\theta\mathrm{d}\theta \\
&=& \int\sqrt{4(1-\sin^2\theta)}\cdot 2\cos\theta\mathrm{d}\theta \\
&=& \int 2\cos\theta\cdot2\cos\theta\mathrm{d}\theta\\
&=& \int 4\cos^2\theta\mathrm{d}\theta.
\end{eqnarray*}

Now use 
\[
\cos^2\theta = \frac{1+\cos(2\theta)}{2}
\]
to finish integrating.