Answer
\[ = {\left( {\tan x} \right)^{10}}\, + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {10{{\tan }^9}x{{\sec }^2}x\,dx} \hfill \\
\hfill \\
set\,\,\,\tan \,\,x = t\,\,\,\,\,then\,\,\,\,\,{\sec ^2}xdx = dt \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
\int_{}^{} {10{{\tan }^9}x{{\sec }^2}x\,dx} = \int_{}^{} {10\,\,{t^9}dt} \hfill \\
\hfill \\
integrate \hfill \\
\hfill \\
= 10\frac{{{t^{10}}}}{{10}}\, + C \hfill \\
\hfill \\
= {t^{10}}\, + C \hfill \\
substitute\,\,back \hfill \\
\hfill \\
= {\left( {\tan x} \right)^{10}}\, + C \hfill \\
\end{gathered} \]
