Answer
$$ - \frac{1}{4}\ln \left| {\cos 4x} \right| - \frac{{{{\tan }^2}4x}}{8} + C$$
Work Step by Step
$$\eqalign{
& \int {{{\tan }^3}4xdx} \cr
& {\text{substitute }}u = 4x,{\text{ }}du = 4dx \cr
& = \frac{1}{4}\int {{{\tan }^3}udu} \cr
& {\text{split off }}\tan \theta \cr
& = \frac{1}{4}\int {{{\tan }^2}u\tan udu} \cr
& {\text{pythagorean identity}} \cr
& = \frac{1}{4}\int {\left( {1 - {{\sec }^2}u} \right)} \tan udu \cr
& {\text{Split }} \cr
& = \frac{1}{4}\int {\tan u} du - \frac{1}{4}\int {{{\sec }^2}u} \tan udu \cr
& {\text{evaluate the integral}} \cr
& = - \frac{1}{4}\ln \left| {\cos u} \right| - \frac{{{{\tan }^2}u}}{8} + C \cr
& {\text{replace }}u = 4x \cr
& = - \frac{1}{4}\ln \left| {\cos 4x} \right| - \frac{{{{\tan }^2}4x}}{8} + C \cr} $$
