Evaluate the following integrals:

Let $\sin x=t$

$d t=\cos x d x$

therefore, $\int \frac{\cos x}{\sqrt{4-\sin ^{2} x}} d x=\int \frac{d t}{\sqrt{2^{2}-t^{2}}}$

Since we have, $\int \frac{1}{\sqrt{a^{2}-x^{2}}} d x=\sin ^{-1}\left(\frac{x}{a}\right)+c$

$=\int \frac{d t}{\sqrt{2^{2}-t^{2}}}=\sin ^{-1}\left(\frac{t}{2}\right)+c=\sin ^{-1}\left(\frac{\sin x}{2}\right)+c$