Answer
$\frac{2}{3}\tan^{\frac{3}{2}}{x}+C$
Work Step by Step
$\int\sqrt{\tan{x}}\sec^2{x}dx$
Let $u=\tan{x}\hspace{8mm}du=\sec^2{x}dx\hspace{8mm}dx=\frac{du}{\sec^2{x}}$
Then
$\int\sqrt{\tan{x}}\sec^2{x}dx=\int\sqrt{u}\sec^2{x}\frac{du}{\sec^2{x}}=\int\sqrt{u}du=\frac{2}{3}u^{\frac{3}{2}}+C=\frac{2}{3}\tan^{\frac{3}{2}}{x}+C$
