## University Calculus: Early Transcendentals (3rd Edition)

$c=2 \tan \sqrt y-x$
Given: $\dfrac{dy}{dx}=\sqrt y \cos^2 \sqrt y$ $\dfrac{dy}{dx}=\dfrac{3x^2}{e^y}$ or, $e^y dy=3x^2 dx$ Re-arrange the given equation and integrate as follows:. Then $\int dx= \int \dfrac{2}{2 \sqrt y \cos^2 \sqrt y} dy$ ....(1) Let us take the help of $u$ substitution. plug $u=\sqrt y \implies \dfrac{dy}{2 \sqrt y}=du$ Equation (1) becomes: $\int \dfrac{2}{ \cos^2 u} du=x+c$ $\implies 2 \tan u=x+c$ Hence, $2 \tan \sqrt y=x+c$ $c=2 \tan \sqrt y-x$