Lol, and I found a substitution instead:
(sin(x) - cos(x))(sin(x) + cos(x)) = sin²(x) - cos²(x)
= sin²(x) - (1 - sin²(x))
= 2 sin²(x) - 1
sin²(x) = (sin(x) - cos(x))(sin(x) + cos(x))/2 + 1/2
int[sin²(x) dx] = int[(sin(x) - cos(x))(sin(x) + cos(x))/2 dx] + int[1/2 dx]
Let u = sin(x) - cos(x), du = (cos(x) + sin(x)) dx; then we have
int[u/2 du] + int[1/2 dx]
= u²/4 + x/2 + C
= (sin(x) - cos(x))²/4 + x/2 + C
= (sin²(x) - 2 sin(x)cos(x) + cos²(x))/4 + x/2 + C
= x/2 + 1/4 - sin(x)cos(x)/2 + C