 Solution

Use $u=\cos x$, so $du=-\sin xdx$. Up to an overall minus sign, the differential form becomesSeeking a monomial integrating factor $u^{\alpha }{y}^{\beta }$, equality of mixed derivatives givesThese give $\alpha =1,\beta =0$, so the integrating factor in the original variables isOne may verify exactness directly without relying on invertibility of the cosine substitution:Thus the implicit first integral isThe equivalence with the differential equation holds directly where $\cos x\ne 0$. The zero solution $y=0$ is included at $C=0$. At a zero of $\cos x$, the original graph equation requires $y=0$ for any finite differentiable continuation, so no extra finite graph with nonzero $y$ is created by multiplying by the vanishing factor. If one allows vertical parametrized integral curves of the differential form, $x=\pi /2+n\pi$ is also such a curve and is already contained in the level set $C=0$.