Answer
$$y = \frac{2}{7}{\tan ^{7/2}}x + \frac{2}{3}{\tan ^{3/2}}x + C$$
Work Step by Step
$$\eqalign{
& y' = \sqrt {\tan x} {\sec ^4}x \cr
& {\text{Write }}y'{\text{ as }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = \sqrt {\tan x} {\sec ^4}x \cr
& {\text{Separate the variables}} \cr
& dy = \sqrt {\tan x} {\sec ^4}xdx \cr
& {\text{Integrate both sides}} \cr
& \int {dy} = \int {\sqrt {\tan x} {{\sec }^4}x} dx \cr
& y = \int {\sqrt {\tan x} {{\sec }^2}x{{\sec }^2}x} dx \cr
& y = \int {\sqrt {\tan x} \left( {{{\tan }^2}x + 1} \right){{\sec }^2}x} dx \cr
& y = \int {\left( {{{\tan }^{5/2}}x + {{\tan }^{1/2}}x} \right){{\sec }^2}x} dx \cr
& y = \frac{{{{\tan }^{7/2}}x}}{{7/2}} + \frac{{{{\tan }^{3/2}}x}}{{3/2}} + C \cr
& y = \frac{2}{7}{\tan ^{7/2}}x + \frac{2}{3}{\tan ^{3/2}}x + C \cr} $$
