$\displaystyle \displaystyle \frac{dy}{dx}=\cos(x+y)$
How can this be solved?
let $\displaystyle u = x+y $
then $\displaystyle y= u-x$ and $\displaystyle \frac{dy}{dx} = \frac{du}{dx} -1 $
so $\displaystyle \frac{du}{dx} -1 = \cos u $
$\displaystyle \frac{du}{1+\cos u} = dx $
integrate both sides
$\displaystyle \tan\Big(\frac{u}{2}\Big) = x + C $ (the substitution $\displaystyle v = \tan\Big(\frac{u}{2}\Big)$ transforms the integral on the left into $\displaystyle \int dv $ )
then $\displaystyle u = 2 \arctan (x+C) $
and finally $\displaystyle y = 2 \arctan(x+C)-x $