= Solution
An <integrating factor for a differential one-form> makes the multiplied form exact. If $dF=\mu g\,dx+\mu f\,dy$, then $F_x=\mu g$ and $F_y=\mu f$. Equality of the mixed <partial derivatives> yields
$$
\boxed{\partial_x(\mu f)=\partial_y(\mu g).}
$$
This is a local compatibility condition under the usual smoothness assumptions.
For the trigonometric equation, where its coefficients are defined and division is legitimate, the slope is $y'=\tan x\tan y$. In the first quadrant it is positive, small near either axis and large near the upper-right corner. The <direction field> below uses normalized line segments, so steep directions remain visible.
Multiplication by $\sin x\cos y$ gives
$$
\cos x\cos y\,dy-\sin x\sin y\,dx=d(\cos x\sin y)=0.
$$
Thus the multiplying function is an <integrating factor>, and the implicit <general solution> is
$$
\boxed{\cos x\sin y=C.}
$$
In the larger square, $\cos x\ge0$. The positive-$C$ curves have positive $y$ and are symmetric about the $y$-axis; the negative-$C$ curves are their reflections. For $0<|C|<1$, a graph representation is
$$
y=\arcsin(C\sec x),\qquad |x|<\arccos|C|,
$$
with endpoints tending to $y=\operatorname{sgn}(C)\pi/2$. The $C=0$ interior curve is $y=0$. The domain boundaries include singular coefficients or zeros of the <integrating factor>, so these endpoint limits and smooth continuations through $x=0$ describe the regularized relation; they must not be mistaken for additional solutions of the literal original equation at undefined points.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2003/ia/paper-2-integrating-factor.png]
{title=Trigonometric direction field, regularized implicit solution family, and the positive higher-order particular solution}
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