This ODE is exact, since
[tex]\dfrac\partial{\partial y}[1-\sin x\tan y]=-\sin x\sec^2y[/tex]
[tex]\dfrac\partial{\partial x}[\cos x\sec^2y]=-\sin x\sec^2y[/tex]
We're looking for a solution of the form [tex]\Psi(x,y)=C[/tex] such that, by the chain rule,
[tex]\dfrac{\mathrm d}{\mathrm dx}\Psi(x,y)=\dfrac{\partial\Psi}{\partial x}\,\mathrm dx+\dfrac{\partial\Psi}{\partial y}\,\mathrm dy=0[/tex]
[tex]\implies\begin{cases}\Psi_x=1-\sin x\tan y\\\Psi_y=\cos x\sec^2y\end{cases}[/tex]
Integrating the first equation with respect to [tex]x[/tex], we have
[tex]\displaystyle\int\Psi_x\,\mathrm dx=\int(1-\sin x\tan y)\,\mathrm dx[/tex]
[tex]\Psi(x,y)=x+\cos x\tan y+f(y)[/tex]
Differentiating with respect to [tex]y[/tex] yields
[tex]\Psi_y=\cos x\sec^2y+f'(y)[/tex]
[tex]\cos x\sec^2y=\cos x\sec^2y+f'(y)[/tex]
[tex]f'(y)=0[/tex]
[tex]\implies f(y)=C[/tex]
So the solution to this ODE is
[tex]\Psi(x,y)=x+\cos x\tan y+C=C[/tex]
or simply
[tex]\Psi(x,y)=x+\cos x\tan y=C[/tex]
