## Calculus: Early Transcendentals (2nd Edition)

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