Find the integral of the function $\tan^{3} 2x \sec 2x$.

(N/A) We want to evaluate $I = \int \tan^{3} 2x \sec 2x \, dx$.
First,rewrite the integrand:
$\tan^{3} 2x \sec 2x = \tan^{2} 2x \cdot \tan 2x \sec 2x = (\sec^{2} 2x - 1) \tan 2x \sec 2x$.
Substitute this into the integral:
$I = \int (\sec^{2} 2x - 1) \tan 2x \sec 2x \, dx = \int \sec^{2} 2x \tan 2x \sec 2x \, dx - \int \tan 2x \sec 2x \, dx$.
For the first part,let $u = \sec 2x$. Then $du = 2 \sec 2x \tan 2x \, dx$,which implies $\sec 2x \tan 2x \, dx = \frac{1}{2} du$.
Thus,$\int \sec^{2} 2x \tan 2x \sec 2x \, dx = \int u^{2} \cdot \frac{1}{2} du = \frac{1}{2} \cdot \frac{u^{3}}{3} = \frac{u^{3}}{6} = \frac{\sec^{3} 2x}{6}$.
For the second part,$\int \tan 2x \sec 2x \, dx = \frac{\sec 2x}{2}$.
Combining these,we get:
$I = \frac{\sec^{3} 2x}{6} - \frac{\sec 2x}{2} + C$,where $C$ is the constant of integration.