Example 4.3. Find the general integral of the equation

$$
\cos ^{2} y \operatorname{ctg} x d x+\sin ^{2} x \operatorname{tg} y d y=0
$$
Solution. Let's separate the variables in the given equation by dividing both sides by the expression $\cos ^{2} y \cdot \sin ^{2} x$:

$$
\frac{\operatorname{ctg} x}{\sin ^{2} x} d x+\frac{\operatorname{tg} y}{\cos ^{2} y} d y=0
$$

Integrating both sides of the equation, we get

$$
\int \frac{\operatorname{ctg} x}{\sin ^{2} x} d x+\int \frac{\operatorname{tg} y}{\cos ^{2} y} d y=C
$$

from which

$$
\frac{-\operatorname{ctg}^{2} x}{2}+\frac{\operatorname{tg}^{2} y}{2}=C
$$

Using the fact that $C$ is an arbitrary constant, we can replace $C$ with $\frac{C}{2}$. Then

$$
\operatorname{tg}^{2} y-\operatorname{ctg}^{2} x=\mathrm{C} .
$$

This is the general integral of the given equation.