write sin2x as
( sinx + cosx ) ^2 = 1 + sin2x
sin2x = ( sinx + cosx ) ^2 - 1
\int \frac{cosx - sinx }{\sqrt{8 - [ ( sinx + cosx ) ^2 -1 )] }}
\int \frac{cosx - sinx }{\sqrt{9 - ( sinx + cosx ) ^2 }}
put sinx + cosx = t
cosx - sinx dx = dt
\int \frac{dt }{\sqrt{9 - ( t ) ^2 }} \rightarrow \sin^{-1}\left( \frac{t}{3}\right)