Answer
$\frac{2}{3}(4-\cos\theta)^{3/2}+c.$
Work Step by Step
Let $u=\cos\theta$, then $du=-\sin\theta d\theta$; we have
\begin{align*}
\int\sin\theta\sqrt{4-\cos\theta}d\theta &= -\int \sqrt{4- u}du \\
&=\frac{1}{3/2}(4-u)^{3/2}+c\\
&=\frac{2}{3}(4-\cos\theta)^{3/2}+c.
\end{align*}
