$\frac{\sin ^{8} x-\cos ^{8} x}{1-2 \sin ^{2} x \cos ^{2} x}$
$\frac{\sin ^{8} x-\cos ^{8} x}{1-2 \sin ^{2} x \cos ^{2} x}=\frac{\left(\sin ^{4} x+\cos ^{4} x\right)\left(\sin ^{4} x-\cos ^{4} x\right)}{\sin ^{2} x+\cos ^{2} x-\sin ^{2} x \cos ^{2} x-\sin ^{2} x \cos ^{2} x}$
$=\frac{\left(\sin ^{4} x+\cos ^{4} x\right)\left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{2} x-\cos ^{2} x\right)}{\left(\sin ^{2} x-\sin ^{2} x \cos ^{2} x\right)+\left(\cos ^{2} x-\sin ^{2} x \cos ^{2} x\right)}$
$=\frac{\left(\sin ^{4} x+\cos ^{4} x\right)\left(\sin ^{2} x-\cos ^{2} x\right)}{\sin ^{2} x\left(1-\cos ^{2} x\right)+\cos ^{2} x\left(1-\sin ^{2} x\right)}$
$=\frac{-\left(\sin ^{4} x+\cos ^{4} x\right)\left(\cos ^{2} x-\sin ^{2} x\right)}{\left(\sin ^{4} x+\cos ^{4} x\right)}$
$=-\cos 2 x$
$\therefore \int \frac{\sin ^{8} x-\cos ^{8} x}{1-2 \sin ^{2} x \cos ^{2} x} d x=\int-\cos 2 x d x=-\frac{\sin 2 x}{2}+\mathrm{C}$