The differential equation \(\cos (x+y) \mathrm{d} y=\mathrm{d} x\) has the general solution given by

\(\cos (x+y) \mathrm{d} y=\mathrm{d} x \)
\( \therefore \frac{\mathrm{d} x}{\mathrm{~d} y}=\cos (x+y)\)
Put \(x+y=\mathrm{u}\) Differentiating w.r.t. \(y\), we get \(\frac{\mathrm{d} x}{\mathrm{~d} y}+1=\frac{\mathrm{du}}{\mathrm{d} y}\)
\(
\therefore \frac{\mathrm{d} x}{\mathrm{~d} y}=\frac{\mathrm{du}}{\mathrm{d} y}-1
\)
Substituting (ii) and (iii) in (i), we get
\( \frac{\mathrm{du}}{\mathrm{d} y}-1=\cos \mathrm{u} \)
\( \therefore \frac{\mathrm{du}}{1+\operatorname{cosu}}=\mathrm{d} y \)
\( \therefore \frac{\mathrm{du}}{2 \cos ^2\left(\frac{\mathrm{u}}{2}\right)}=\mathrm{d} y\)
Integrating on both sides, we get
\(\frac{1}{2} \int \sec ^2\left(\frac{\mathrm{u}}{2}\right) \mathrm{du}=\int \mathrm{d} y \)
\(\therefore y=\tan \left(\frac{x+y}{2}\right)+\mathrm{c}\)