$\int {{{\tan }^2}} x{\sec ^4}xdx$

Let $I = \int {{{\tan }^2}} x{\sec ^4}xdx$

Let's put $\tan x = t \Rightarrow {\sec ^2}xdx = dt$

therefore,$I = \int {{t^2}} \left( {1 + {t^2}} \right)dt = \int {\left( {{t^2} + {t^4}} \right)} dt$

$= \frac{{{t^3}}}{3} + \frac{{{t^5}}}{5} + C = \frac{{{{\tan }^5}x}}{5} + \frac{{{{\tan }^3}x}}{3} + C$