Answer
\[ = \tan x - x + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {{{\tan }^2}xdx} \hfill \\
\hfill \\
write\,\,\,{\tan ^2}x = {\sec ^2}x - 1 \hfill \\
\hfill \\
= \int_{}^{} {\,\left( {{{\sec }^2}x - 1} \right)} \,\,dx \hfill \\
\hfill \\
{\text{split}}\,\,{\text{the}}\,\,{\text{integrand}} \hfill \\
\hfill \\
= \int_{}^{} {{{\sec }^2}xdx\, - \int_{}^{} {dx} } \hfill \\
\hfill \\
integrate \hfill \\
\hfill \\
= \tan x - x + C \hfill \\
\end{gathered} \]
