Evaluate the following integrals:

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

$\tan ^{2} 2 x=\sec ^{2} 2 x-1$

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

$\Rightarrow \sec ^{2} 2 x \tan 2 x \cdot \sec 2 x d x-\tan 2 x \cdot \sec 2 x d x$

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

$\Rightarrow \int \sec ^{2} 2 x \cdot \tan 2 x \cdot \sec 2 x \cdot d x-\frac{\sec 2 x}{2}+c$

Assume sec2x $=\mathrm{t}$

$d(\sec 2 x)=d t$

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

$\Rightarrow \int \mathrm{t}^{2} \cdot \mathrm{dt}-\frac{\sec 2 \mathrm{x}}{2}+\mathrm{c}$

$\Rightarrow \frac{t^{3}}{3}-\frac{\sec 2 x}{2}+c$

But $t=\sec 2 x$

$\Rightarrow \frac{(\sec 2 x)^{2}}{3}-\frac{\sec 2 x}{2}+c$