Answer
\[ = 6\tan x + 2{\tan ^3}x + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {6\,{{\sec }^4}xdx} \hfill \\
\hfill \\
write\,{\text{ }}{\sec ^4}x{\text{ as }}\,{\sec ^2}x{\sec ^2}x \hfill \\
\hfill \\
= 6\int_{}^{} {{{\sec }^2}x{{\sec }^2}xdx} \hfill \\
\hfill \\
use\,\,\,{\sec ^2}x = 1 + {\tan ^2}x \hfill \\
\hfill \\
then \hfill \\
\hfill \\
= 6\int_{}^{} {\,\left( {1 + {{\tan }^2}x} \right){{\sec }^2}xdx} \hfill \\
\hfill \\
= 6\int_{}^{} {{{\sec }^2}xdx + 6\int_{}^{} {{{\tan }^2}x{{\sec }^2}xdx} } \hfill \\
\hfill \\
integrate \hfill \\
\hfill \\
= 6\tan x + 2{\tan ^3}x + C \hfill \\
\end{gathered} \]