First Order ODE/y' = sin^2 (x - y + 1)

Theorem
The first order ODE:
 * $\dfrac {\d y} {\d x} = \map {\sin^2} {x - y + 1}^2$

has the general solution:
 * $\map \tan {x - y + 1} = x + C$

Proof
Make the substitution:
 * $z = x - y + 1$

Then from First Order ODE in form $y' = f (a x + b y + c)$ with $a = 1, b = - 1$: