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

$\tan y=-\cos x-x \sin x+c$
In order to to solve the given differential equation we will have to separate the variables and then integrate. Here, we have $\sec x dy+x \cos^2 y dx=0$ or, $\int \dfrac{dy}{\cos^2 y}=-\int \dfrac{x}{\sec x} dx$ or, $x \sin x-\int (-\sin x) dx =\tan y$ Then, we have $-[x \cos x+\cos x +c]=\tan y$ Hence, $\tan y=-\cos x-x \sin x+c$