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

Theorem
The first order ODE:
 * $\dfrac {\mathrm d y} {\mathrm d x} = \sin^2 \left({x - y + 1}\right)^2$

has the solution:
 * $\tan \left({x - y + 1}\right) = x + C$

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

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