$dy/dx = \sin (x + y)$
$dy = \sin (x + y)~dx$
Let
$z = x + y$
$dz = dx + dy$
$dy = dz - dx$
$dz - dx = \sin z~dx$
$dz = \sin z~dx + dx$
$dz = (\sin z + 1)~dx$
$\dfrac{dz}{\sin z + 1} = dx$
$\dfrac{dz}{1 + \sin z} \cdot \dfrac{1 - \sin z}{1 - \sin z} = dx$
$\dfrac{(1 - \sin z)~dz}{1 - \sin^2 z} = dx$
$\dfrac{(1 - \sin z)~dz}{\cos^2 z} = dx$
$\dfrac{~dz}{\cos^2 z} - \dfrac{\sin z~dz}{\cos^2 z} = dx$
$\sec^2 z~dz - \cos^{-2} z \sin z~dz = dx$
$\displaystyle \int \sec^2 z~dz + \int \cos^{-2} z ~ (-\sin z~dz) = \int dx$
$\tan z - \cos^{-1} z = x + c$
$\tan z - \dfrac{1}{\cos z} = x + c$
$\tan z - \sec z = x + c$
$\tan (x + y) - \sec (x + y) = x + c$
$ x + c = \tan (x + y) - \sec (x + y)$ answer
