Answer
$$-\frac{1}{6} (\cos 4 x+1)^{3 / 2}+C $$
Work Step by Step
Given $$ \int \sin 4 x \sqrt{\cos 4 x+1} d x $$
Let
$$u=\cos 4 x+1 \ \ \ \ \Rightarrow \ \ \ du =-4\sin 4x dx $$
\begin{aligned} \int \sin 4 x \sqrt{\cos 4 x+1} d x &=-\frac{1}{4} \int \sqrt{u} d u \\ &=-\frac{1}{4} \int u^{1 / 2} d u \\ &=-\frac{1}{4}\left(\frac{2}{3} u^{3 / 2}+C\right) \\ &=-\frac{1}{6} u^{3 / 2}+C\\
&=-\frac{1}{6} (\cos 4 x+1)^{3 / 2}+C \end{aligned}
