Answer
\[ = \frac{2}{3}\,{\left( {\tan x} \right)^{\frac{3}{2}}} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {{{\sec }^2}x{{\tan }^{\frac{1}{2}}}xdx} \hfill \\
\hfill \\
set\,\,\,u = \tan x\,\,\,\,\,then{\text{ }}du = \,\,{\sec ^2}xdx \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
\int_{}^{} {{{\sec }^2}x{{\tan }^{\frac{1}{2}}}xdx} = \int_{}^{} {{u^{\frac{1}{2}}}du} \hfill \\
\hfill \\
integrate \hfill \\
\hfill \\
= \frac{2}{3}{u^{\frac{3}{2}}} + C \hfill \\
\hfill \\
substituting\,\,back\,\,u\, = \,\tan x \hfill \\
\hfill \\
= \frac{2}{3}\,{\left( {\tan x} \right)^{\frac{3}{2}}} + C \hfill \\
\hfill \\
\end{gathered} \]
