## Thomas' Calculus 13th Edition

$\tan y=-\cos x-x \sin x+c$
Re-arrange the given equation as: $\sec x dy+x \cos^2 y dx=0$ Now, $\int \dfrac{1}{\cos^2 y}dy=-\int \dfrac{xdx}{\sec x}$ $\implies (x) (\sin x)-\int (-\sin x) dx =\tan (y)$ This gives: $-[x \cos x+\cos x +c]=\tan (y)$ Hence, $\tan y=-\cos x-x \sin x+c$