Answer
\[ = \frac{{{{\sec }^3}x}}{3} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\tan x{{\sec }^3}xdx} \hfill \\
\hfill \\
{\text{rewrite}}\,\,\,{\text{the}}\,\,{\text{integrand}} \hfill \\
\hfill \\
\int_{}^{} {{{\sec }^2}x\tan x\sec xdx} \hfill \\
\hfill \\
use\,\,\,{\sec ^2}x = {u^2}\,\,\,\,\,then\,\,\,\tan x\sec x\,dx = \,du \hfill \\
\hfill \\
\int_{}^{} {{{\sec }^2}x\tan x\sec xdx} = \int_{}^{} {{u^2}du} \hfill \\
\hfill \\
integrate \hfill \\
\hfill \\
= \frac{{{u^3}}}{3} + C \hfill \\
\hfill \\
substituting\,\,back\,\,u\, = \,\sec x \hfill \\
\hfill \\
= \frac{{{{\sec }^3}x}}{3} + C \hfill \\
\end{gathered} \]
