The solution of $\cos (x+y) d y=d x$, is

$y=\tan \left(\frac{x+y}{2}\right)+C$

We have, $\cos (x+y) d y=d x$

Putting $x+y=v$ and $1+\frac{d y}{d x}=\frac{d v}{d x}$, we get

$\cos v\left(\frac{d v}{d x}-1\right)=1$

$\Rightarrow \frac{d v}{d x}=\frac{1+\cos v}{\cos v}$

$\Rightarrow \frac{\cos v}{1+\cos v} d v=d x$

$\Rightarrow \left(1-\frac{1}{1+\cos v}\right) d v=d x$

$\Rightarrow \int\left(1-\frac{1}{2} \sec ^2 \frac{v}{2}\right) d v=\int d x$

$\Rightarrow v-\tan \frac{v}{2}=x+C$

$\Rightarrow x+y-\tan \left(\frac{x+y}{2}\right)=x+C \Rightarrow y=\tan \frac{x+y}{2}+C$