Answer
$$\frac{2 \tan ^{\frac{3}{2}} x}{3}+C$$
Work Step by Step
Given
$$ \int \sqrt{\tan x} \sec ^{2} x d x$$
Let $ u=\tan x\to du=\sec^2 xdx$
\begin{aligned}
\int \sqrt{\tan x} \sec ^{2} x d x &=\int \sqrt{u} d u \\ &=\frac{u^{\frac{3}{2}}}{\frac{3}{2}}+C \\ &=\frac{2 \tan ^{\frac{3}{2}} x}{3}+C
\end{aligned}
