Prove

$\tan ^{3} 2 x \sec 2 x=\tan ^{2} 2 x \tan 2 x \sec 2 x$

$=\left(\sec ^{2} 2 x-1\right) \tan 2 x \sec 2 x$

$=\sec ^{2} 2 x \cdot \tan 2 x \sec 2 x-\tan 2 x \sec 2 x$

$\therefore \int \tan ^{3} 2 x \sec 2 x d x=\int \sec ^{2} 2 x \tan 2 x \sec 2 x d x-\int \tan 2 x \sec 2 x d x$

$=\int \sec ^{2} 2 x \tan 2 x \sec 2 x d x-\frac{\sec 2 x}{2}+\mathrm{C}$

Let $\sec 2 x=t$

$\therefore 2 \sec 2 x \tan 2 x d x=d t$

$\therefore \int \tan ^{3} 2 x \sec 2 x d x=\frac{1}{2} \int t^{2} d t-\frac{\sec 2 x}{2}+\mathrm{C}$

$=\frac{t^{3}}{6}-\frac{\sec 2 x}{2}+\mathrm{C}$

$=\frac{(\sec 2 x)^{3}}{6}-\frac{\sec 2 x}{2}+\mathrm{C}$