Given differential equation is
\(\begin{array}{rlrl}
\Rightarrow \quad & & \sin y \frac{d y}{d x} & =\cos y(1-x \cos y) \\
\Rightarrow & \quad \sin y \frac{d y}{d x} & =\cos y-x \cos ^2 y \\
\Rightarrow \quad & \quad \frac{\sin y}{\cos ^2 y} \frac{d y}{d x} & =\frac{1}{\cos y}-x \\
\Rightarrow \quad & \sec y \tan y \frac{d y}{d x} & =\sec y-x
\end{array}\)
Let \(\quad \sec y=t\)
\(\begin{aligned}
& \Rightarrow \sec y \tan y \frac{d y}{d x}=\frac{d t}{d x} \\
& \therefore \quad \frac{d t}{d x}=t-x \\
& \Rightarrow \quad \frac{d t}{d x}+(-t)=-x \\
& \therefore \quad \mathrm{IF}=e^{\int-d x}=e^{-x}
\end{aligned}\)
Now, required solution
\(\begin{aligned}
\Rightarrow \quad t(I F) & =\int(-x)(I F) d x+c \\
\sec y\left(e^{-x}\right) & =\int(-x) e^{-x} d x+c \\
& =\left(x e^{-x}\right)-\int e^{-x} d x+c \\
& =x e^{-x}+e^{-x}+c \\
\Rightarrow \quad & \sec y=x+1+c e^x
\end{aligned}\)