Answer
$-\frac{1}{6}(2-\sin4\theta)^{\frac{3}{2}}+C$
Work Step by Step
Let $u=2-\sin4\theta$. Then $du=-4\cos 4\theta d\theta$, so $\cos 4\theta d\theta=-\frac{1}{4}du$.
$\int\cos 4\theta\sqrt{2-\sin 4\theta}d\theta$
$=\int\sqrt{2-\sin 4\theta}\cos 4\theta d\theta$
$=\int\sqrt{u}*(-\frac{1}{4})du$
$=-\frac{1}{4}\int u^{\frac{1}{2}}du$
$=-\frac{1}{4}*\frac{u^{\frac{3}{2}}}{\frac{3}{2}}+C$
$=-\frac{1}{6}u^{\frac{3}{2}}+C$
$=-\frac{1}{6}(2-\sin4\theta)^{\frac{3}{2}}+C$