If $\int\frac{\sin^8x-\cos^8x}{1-2\sin^2x\cos^2x}dx=A\sin 2x+B$, is:

$A=-\frac{1}{2}$

$\int\frac{\sin^8x-\cos^8x}{1-2\sin^2x\cos^2x}dx=\int\frac{(\sin^4x-\cos^4x)(\sin^4x+\cos^4x)}{(1-2\sin^2x\cos^2x)}dx$

$=\int\frac{(\sin^2x-\cos^2x)(\sin^2x+\cos^2x)}{(1-2\sin^2x\cos^2x)}×[(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x]dx$

$=\int\frac{\cos 2x.(1-2\sin^2x\cos^2x)}{(1-2\sin^2x\cos^2x)}dx=-\int\cos 2x\,dx=\frac{-1}{2}\sin 2x+B⇒A=\frac{-1}{2}$