Find $\int \frac{\sin 2x \cos 2x \, dx}{\sqrt{9-\cos^{4}(2x)}}$

(NONE) Let $I = \int \frac{\sin 2x \cos 2x \, dx}{\sqrt{9-\cos^{4}(2x)}}$.
Substitute $t = \cos^{2}(2x)$.
Then,$dt = 2 \cos(2x) \cdot (-\sin(2x)) \cdot 2 \, dx = -4 \sin(2x) \cos(2x) \, dx$.
This implies $\sin(2x) \cos(2x) \, dx = -\frac{1}{4} dt$.
Substituting these into the integral:
$I = \int \frac{-\frac{1}{4} dt}{\sqrt{9-t^{2}}} = -\frac{1}{4} \int \frac{dt}{\sqrt{3^{2}-t^{2}}}$.
Using the standard integral formula $\int \frac{dx}{\sqrt{a^{2}-x^{2}}} = \sin^{-1}(\frac{x}{a}) + C$:
$I = -\frac{1}{4} \sin^{-1}\left(\frac{t}{3}\right) + C$.
Substituting $t = \cos^{2}(2x)$ back:
$I = -\frac{1}{4} \sin^{-1}\left(\frac{\cos^{2}(2x)}{3}\right) + C$.