$\int \tan ^3 2 x \sec 2 x d x=$

$\frac{1}{6}\left[\sec ^3 2 x-3 \sec 2 x\right]$

$I=\int \tan ^2 2 x \tan 2 x \sec 2 x d x=\int\left(\sec ^2 2 x-1\right) \sec 2 x \tan 2 x d x $

Put  $\sec 2 x=t$     ∴ $2 \sec 2 x \tan 2 x dx=dt $

∴   $I=\frac{1}{2} \int\left(t^2-1\right) d t=\frac{1}{2}\left(\frac{t^3}{3}-t\right)=\frac{1}{6}\left(\sec ^3 2 x-3 \sec 2 x\right)$

Hence (2) is the correct answer.