If $\int \frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x} d x=a \sin 2 x+C$, then $a=$

$-\frac{1}{2}$

We have,

$I=\int \frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x} d x$

$\Rightarrow I=\int \frac{\left(\sin ^4 x-\cos ^4 x\right)\left(\sin ^4 x+\cos ^4 x\right)}{\left(\sin ^2 x+\cos ^2 x\right)^2-2 \sin ^2 x \cos ^2 x} d x$

$\Rightarrow I=\int\left(\sin ^2 x-\cos ^2 x\right) d x$

$\Rightarrow I=-\int \cos 2 x d x=-\frac{1}{2} \sin 2 x+C$

$\Rightarrow -\frac{1}{2} \sin 2 x+C=a \sin 2 x+C \Rightarrow a=-\frac{1}{2}$