General solution of the differential equation \(\cos x(1+\cos y) \mathrm{d} x-\sin y(1+\sin x) \mathrm{d} y=0\) is

Given differential equation is
\(\begin{aligned}
& \cos x(1+\cos y) \mathrm{d} x-\sin y(1+\sin x) \mathrm{d} y=0 \\
& \Rightarrow \cos x(1+\cos y) \mathrm{d} x=\sin y(1+\sin x) \mathrm{d} y \\
& \Rightarrow \frac{\cos x}{1+\sin x} \mathrm{~d} x=\frac{\sin y}{1+\cos y} \mathrm{~d} y
\end{aligned}\)
Integrating on both sides, we get
\(\begin{aligned}
& \int \frac{\cos x}{1+\sin x} \mathrm{~d} x=\frac{\sin y}{1+\cos y} \mathrm{~d} y+\log |\mathrm{c}| \\
& \Rightarrow \log |1+\sin x|=-\log |1+\cos y|+\log |\mathrm{c}| \\
& \Rightarrow \log |1+\sin x|+\log |1+\cos y|=\log |\mathrm{c}| \\
& \Rightarrow \log |(1+\sin x)(1+\cos y)|=\log |\mathrm{c}| \\
& \Rightarrow(1+\sin x)(1+\cos y)=\mathrm{c}
\end{aligned}\)