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$(3 + y + 2y^2 \sin^2 x)\,dx
$(3 + y + 2y^2 \sin^2 x)\,dx + (x + 2xy - y\sin 2x)\,dy = 0$
$M = 3 + y + 2y^2 \sin^2 x$
$\dfrac{\partial M}{\partial y} = 1 + 4y \sin^2 x$
$N = x + 2xy - y\sin 2x$
$\begin{aligned}
\dfrac{\partial N}{\partial x} &= 1 + 2y - 2y\cos 2x \\
&= 1 + 2y - 2y\Big[ 1 - 2 \sin^2 x \Big] \\
&= 1 + 2y - 2y + 4y \sin^2 x \\
&= 1 + 4y \sin^2 x
\end{aligned}$
$\dfrac{\partial M}{\partial y} = \dfrac{\partial N}{\partial x}$ thus, exact.
$\partial F = M \, \partial x$
$\begin{aligned}
\partial F &= (3 + y + 2y^2 \sin^2 x)\partial x \\
&= (3 + y)\partial x + 2y^2 \Big[ \frac{1}{2}(1 - \cos 2x) \Big] \partial x \\
&= (3 + y)\partial x + (y^2 - y^2 \cos 2x) \partial x
\end{aligned}$
$F = 3x + xy + xy^2 - \frac{1}{2}y^2 \sin 2x + f(y)$
$\dfrac{\partial F}{\partial y} = x + 2xy - y \sin 2x + f'(y)$
$\dfrac{\partial F}{\partial y} = N$
$x + 2xy - y \sin 2x + f'(y) = x + 2xy - y\sin 2x$
$f'(y) = 0$
$f(y) = 0$
$F = c$
$3x + xy + xy^2 - \frac{1}{2}y^2 \sin 2x = c$ answer