Answer
\[ = \frac{2}{3}{\tan ^{\frac{3}{2}}}x + \frac{2}{7}{\tan ^{\frac{7}{2}}}x + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\sqrt {\tan x} {{\sec }^4}xdx} \hfill \\
\hfill \\
{\text{rewrite}}\,\,{\text{the}}\,\,{\text{integrand}} \hfill \\
\hfill \\
\int_{}^{} {\sqrt {\tan x} \,\sec {\,^2}x{{\sec }^2}xdx} \hfill \\
\hfill \\
use\,\,{\sec ^2}x = 1 + {\tan ^2}x \hfill \\
\hfill \\
= \int_{}^{} {\sqrt {\tan x} } \,\left( {1 + {{\tan }^2}x} \right){\sec ^2}xdx \hfill \\
\hfill \\
use\,\,\tan x = u\,\,\,\, \to \,\,du = {\sec ^2}xdx \hfill \\
\hfill \\
= \int_{}^{} {{u^{\frac{1}{2}}}\,\left( {1 + {u^2}} \right)du} \hfill \\
\hfill \\
multiply \hfill \\
\hfill \\
\int_{}^{} {\,\left( {{u^{\frac{1}{2}}} + {u^{\frac{5}{2}}}} \right)} \,du \hfill \\
\hfill \\
integrate \hfill \\
\hfill \\
= \frac{{{u^{\frac{3}{2}}}}}{{\frac{3}{2}}} + \frac{{{u^{\frac{7}{2}}}}}{{\frac{7}{2}}} + C \hfill \\
\hfill \\
substituting\,\,back\,\,u\, = \,\tan x \hfill \\
\hfill \\
= \frac{2}{3}{\tan ^{\frac{3}{2}}}x + \frac{2}{7}{\tan ^{\frac{7}{2}}}x + C \hfill \\
\end{gathered} \]
