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

.: Let $I = \int {{{\tan }^3}2x\sec 2x} dx$

$I = \int {{{\tan }^2}2x \cdot \tan 2x \cdot \sec 2x} dx = \int {\left( {{{\sec }^2}2x - 1} \right)} \cdot \sec 2x\tan 2x\,dx$

Put $\sec 2x = t$ $\Rightarrow$ $2\sec 2x\tan 2x\,dx = dt$

$\therefore$ $I = \cfrac{1}{2}\int {\left( {{t^2} - 1} \right)dt} = \cfrac{1}{2}\left( {\cfrac{{{t^3}}}{3} - t} \right) + C = \cfrac{1}{6}{\sec ^3}2x - \cfrac{1}{2}\sec 2x + C$