Evaluate the following integrals:
$\int \tan ^{3} 2 x \sec 2 x d x$
$\tan ^{3} 2 x . \sec 2 x=\tan ^{2} 2 x . \tan 2 x . \sec 2 x . d x$
$\tan ^{2} 2 x=\sec ^{2} 2 x-1$
$\Rightarrow \tan ^{2} 2 x \cdot \tan 2 x \cdot \sec 2 x \cdot d x=\left(\sec ^{2} 2 x-1\right) \cdot \tan 2 x \cdot \sec 2 x \cdot d x$
$\Rightarrow \sec ^{2} 2 x \tan 2 x \cdot \sec 2 x d x-\tan 2 x \cdot \sec 2 x d x$
$\therefore \int \sec ^{2} 2 x \cdot \tan 2 x \cdot \sec 2 x d x-\int \tan 2 x \cdot \sec 2 x \cdot d x$
$\Rightarrow \int \sec ^{2} 2 x \cdot \tan 2 x \cdot \sec 2 x \cdot d x-\frac{\sec 2 x}{2}+c$
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 $t=\sec 2 x$
$\Rightarrow \frac{(\sec 2 x)^{2}}{3}-\frac{\sec 2 x}{2}+c$
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