Answer
$y=[\tan^{-1} (\dfrac{x+c}{2})]^2$
Work Step by Step
Given: $dy=\sqrt y \cos^2 \sqrt y dy$
Now separate the differentials and integrate.
$\int dx=\int \dfrac{\sec^2 y dy}{\sqrt y}$
Plug in $a=\sqrt y \implies 2da=\dfrac{1}{\sqrt y}dy$
Then $\int \dfrac{\sec^2 y dy}{\sqrt y}=x+c \implies 2 \int sec^2 (a) da=x+c$
or, $2 \tan a=x+c \implies 2 \tan \sqrt y=x+c$
This implies that $\sqrt y=\tan^{-1} (x+c/2)$
Thus, $y=[\tan^{-1} (\dfrac{x+c}{2})]^2$
