Answer
\[ = \frac{{{{\tan }^{10}}x}}{{10}} + \frac{{{{\tan }^{12}}x}}{{10}} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {{{\tan }^9}x{{\sec }^4}xdx} \hfill \\
\hfill \\
{\text{rewrite}}\,\,{\text{the}}\,\,{\text{integrand}} \hfill \\
\hfill \\
= \int_{}^{} {{{\tan }^9}x \cdot {{\sec }^2}x \cdot {{\sec }^2}xdx} \hfill \\
\hfill \\
use{\text{ }}{\sec ^2}x = 1 + {\tan ^2}x \hfill \\
\hfill \\
= \int_{}^{} {{{\tan }^9}x\,\left( {1 + {{\tan }^2}x} \right){{\sec }^2}xdx} \hfill \\
\hfill \\
use\,\,\,\,\,\tan x = u\,\,\,\,\,then\,\,\,{\sec ^2}xdx = du \hfill \\
\hfill \\
= \int_{}^{} {{u^9}\,\left( {1 + {u^2}} \right)du} \hfill \\
\hfill \\
= \int_{}^{} {\,\left( {{u^9} + {u^{11}}} \right)} du \hfill \\
\hfill \\
= \int_{}^{} {{u^9}du\,\, + \int_{}^{} {{u^{11}}} } du \hfill \\
\hfill \\
integrate \hfill \\
\hfill \\
= \frac{{{u^{10}}}}{{10}} + \frac{{{u^{12}}}}{{12}} + C \hfill \\
\hfill \\
substituting\,\,back\,\,u = \tan \,x \hfill \\
\hfill \\
= \frac{{{{\tan }^{10}}x}}{{10}} + \frac{{{{\tan }^{12}}x}}{{10}} + C \hfill \\
\end{gathered} \]
