Answer
$$\frac{{{{\sec }^4}4x}}{{16}} + C$$
Work Step by Step
$$\eqalign{
& \int {\tan 4x{{\sec }^4}4x} dx \cr
& {\text{split }}{\sec ^4}x \cr
& = \int {\tan 4x\sec 4x{{\sec }^3}4x} dx \cr
& = \int {{{\sec }^3}4x\sec 4x\tan 4x} dx \cr
& {\text{substitute }}u = \sec 4x,{\text{ }}du = \sec 4x\tan 4x\left( 4 \right)dx \cr
& \frac{1}{4}du = \sec 4x\tan 4xdx \cr
& = \int {{u^3}\left( {\frac{1}{4}du} \right)} \cr
& = \frac{1}{4}\int {{u^3}du} \cr
& {\text{find the antiderivative by the power rule}} \cr
& = \frac{{{u^4}}}{{16}} + C \cr
& {\text{write in terms of }}x,{\text{ replace }}u = \tan x \cr
& = \frac{{{{\sec }^4}4x}}{{16}} + C \cr} $$
