Answer
\[ = \frac{{{{\sec }^{12}}x}}{{12}} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {{{\sec }^{12}}x\tan xdx} \hfill \\
\hfill \\
split\,\,\,\,\,{\sec ^{12}}x = {\sec ^{11}}x\sec x \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
\int_{}^{} {{{\sec }^{12}}\tan x\,dx} = \int_{}^{} {{{\sec }^{11}}x\sec x\tan xdx} \hfill \\
\hfill \\
set\,\,u = \sec \,x \to \,du = \sec x\tan xdx \hfill \\
\hfill \\
then \hfill \\
\hfill \\
\int_{}^{} {{{\sec }^{11}}x\sec x\tan xdx} = \int {{u^{11}}du} \hfill \\
\hfill \\
integrate \hfill \\
\hfill \\
\frac{{{u^{12}}}}{{12}} + C \hfill \\
\hfill \\
substitute\,\,back\,\,u = \sec x \hfill \\
\hfill \\
= \frac{{{{\sec }^{12}}x}}{{12}} + C \hfill \\
\end{gathered} \]
