Answer
$\frac{1}{9}{\tan ^9}x + \frac{2}{7}{\tan ^7}x + \frac{1}{5}{\tan ^5}x + C$
Work Step by Step
$$\eqalign{
& \text{Let }I= \int {{{\tan }^4}x{{\sec }^6}x} dx \cr
& {\text{Split }}{\sec ^6}x{\text{ as }}{\sec ^4}x{\sec ^2}x \cr
& I = \int {{{\tan }^4}x{{\sec }^4}x{{\sec }^2}x} dx \cr
& {\text{Use the property }}{\left( {{a^m}} \right)^n} = {a^{mn}} \cr
& = \int {{{\tan }^4}x{{\left( {{{\sec }^2}x} \right)}^2}{{\sec }^2}x} dx \cr
& {\text{Rewrite }}{\sec ^2}x{\text{ using }}{\tan ^2}\theta = {\sec ^2}\theta - 1 \cr
& I = \int {{{\tan }^4}x{{\left( {{{\tan }^2}x + 1} \right)}^2}{{\sec }^2}x} dx \cr
& {\text{Expand the binomial}} \cr
& I= \int {{{\tan }^4}x\left( {{{\tan }^4}x + 2{{\tan }^2}x + 1} \right){{\sec }^2}x} dx \cr
& = \int {\left( {{{\tan }^8}x + 2{{\tan }^6}x + {{\tan }^4}x} \right){{\sec }^2}x} dx \cr
& {\text{Let }}u = \tan x,{\text{ }}du = {\sec ^2}xdx \cr
& {\text{Substituting}} \cr
& I= \int {\left( {{u^8} + 2{u^6} + {u^4}} \right)du} \cr
& {\text{Use the power rule }}\int {{u^n}} du = \frac{1}{{n + 1}}{u^{n + 1}} + C \cr
& I= \frac{1}{9}{u^9} + \frac{2}{7}{u^7} + \frac{1}{5}{u^5} + C \cr
& {\text{Write in terms of }}x,{\text{ substitute }}\tan x{\text{ for }}u \cr
& I= \frac{1}{9}{\tan ^9}x + \frac{2}{7}{\tan ^7}x + \frac{1}{5}{\tan ^5}x + C \cr} $$
