$\begin{align}  & \int{{{\tan }^{5}}x{{\sec }^{2}}xdx}=\int{{{\tan }^{4}}x{{\sec }^{2}}x\tan xdx} \\ & \text{Let }u=\tan x \\ & \frac{du}{dx}={{\sec }^{3}}x,\text{ }dx=\frac{du}{{{\sec }^{2}}x} \\ & \therefore \int{{{\tan }^{5}}{{\sec }^{2}}xdx}=\int{{{u}^{4}}{{\sec }^{2}}x\cdot u}\cdot \frac{du}{{{\sec }^{2}}x} \\ & \int{{{\tan }^{5}}{{\sec }^{2}}xdx}=\int{{{u}^{5}}du}=\frac{{{u}^{6}}}{6}+C=\frac{1}{6}{{\tan }^{6}}x+C \\\end{align}$