Problem 05 | Substitution Suggested by the Equation

$dy/dx = \sin (x + y)$

$dy = \sin (x + y)~dx$
 

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