Answer
$$\frac{{{{\tan }^7}x}}{7} + \frac{{{{\tan }^5}x}}{5} + C$$
Work Step by Step
$$\eqalign{
& \int {{{\tan }^4}\theta {{\sec }^4}\theta } d\theta \cr
& {\text{split }}{\sec ^4}\theta \cr
& = \int {{{\tan }^4}\theta {{\sec }^2}\theta {{\sec }^2}\theta } d\theta \cr
& {\text{identity se}}{{\text{c}}^2}\theta = {\tan ^2}\theta + 1 \cr
& = \int {{{\tan }^4}\theta \left( {{{\tan }^2}\theta + 1} \right){{\sec }^2}} \theta d\theta \cr
& = \int {\left( {{{\tan }^6}\theta + {{\tan }^4}\theta } \right){{\sec }^2}\theta } d\theta \cr
& {\text{substitute }}u = \tan x,{\text{ }}du = {\sec ^2}xdx \cr
& = \int {\left( {{u^6} + {u^4}} \right)du} \cr
& {\text{find the antiderivative by the power rule}} \cr
& = \frac{{{u^7}}}{7} + \frac{{{u^5}}}{5} + C \cr
& {\text{write in terms of }}x,{\text{ replace }}u = \tan x \cr
& = \frac{{{{\tan }^7}x}}{7} + \frac{{{{\tan }^5}x}}{5} + C \cr} $$
