$\begin{align}  & \int{{{\tan }^{3}}x{{\sec }^{2}}xdx}=\int{{{(\tan x)}^{3}}{{\sec }^{2}}xdx} \\ & \text{Let }u=\tan x,\text{  }\frac{du}{dx}={{\sec }^{2}}x,\text{ }dx=\frac{du}{{{\sec }^{2}}x} \\ & \int{{{\tan }^{3}}x{{\sec }^{2}}x}=\int{{{u}^{3}}{{\sec }^{2}}x(\tfrac{dx}{{{\sec }^{2}}x})}=\int{{{u}^{3}}dx} \\ & \int{{{\tan }^{3}}{{\sec }^{2}}x}=\frac{{{u}^{4}}}{4}+C=\frac{{{\tan }^{4}}x}{4}+C \\\end{align}$