Answer
$\tan^2(2x)-2\ln|\sec2x|+C$
Work Step by Step
Use the Reduction Formula:
$\int\tan^naxdx=\dfrac{\tan^{n-1}ax}{a(n-1)}-\int \tan^{n-2}axdx$ and $\int\tan axdx=\dfrac{1}{a}\ln|\sec ax|+C$
Let $I=\int 4\tan^3 2xdx\\=4[\dfrac{\tan^22x}{4}-\int\tan2xdx ]\\ =\tan^2(2x)-2\ln|\sec2x|+C$
