Answer
$$\frac{1}{8}{\tan ^2}4x + \frac{1}{4}\ln \left| {\cos 4x} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {{{\tan }^3}4x} dx \cr
& {\text{split exponent of }}{\tan ^3}4x \cr
& = \int {{{\tan }^2}4x} \tan 4xdx \cr
& {\text{pythagorean identity 1 + ta}}{{\text{n}}^2}\theta = {\sec ^2}\theta \cr
& = \int {\left( {{{\sec }^2}4x - 1} \right)} \tan 4xdx \cr
& = \int {\left( {\tan 4x{{\sec }^2}4x - \tan 4x} \right)} dx \cr
& {\text{sum rule}} \cr
& = \int {\tan 4x{{\sec }^2}4x} dx - \int {\tan 4x} dx \cr
& {\text{find the antiderivatives}} \cr
& = \frac{1}{{4\left( 2 \right)}}{\tan ^2}4x + \frac{1}{4}\ln \left| {\cos 4x} \right| + C \cr
& = \frac{1}{8}{\tan ^2}4x + \frac{1}{4}\ln \left| {\cos 4x} \right| + C \cr} $$