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$\int \frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x} d x=$
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$\frac{-1}{2} \sin 2 x+c$
$\int \frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x} d x$
$\begin{aligned} & =\int \frac{\left(\sin ^4 x+\cos ^4 x\right)\left(\sin ^2 x+\cos ^2 x\right)\left(\sin ^2 x-\cos ^2 x\right)}{\left(\sin ^4 x+\cos ^4 x\right)} d x \\ & =\int\left(\sin ^2 x-\cos ^2 x\right) d x=-\int \cos 2 x d x \\ & =-\frac{1}{2} \sin 2 x+C .\end{aligned}$
$\begin{aligned} & =\int \frac{\left(\sin ^4 x+\cos ^4 x\right)\left(\sin ^2 x+\cos ^2 x\right)\left(\sin ^2 x-\cos ^2 x\right)}{\left(\sin ^4 x+\cos ^4 x\right)} d x \\ & =\int\left(\sin ^2 x-\cos ^2 x\right) d x=-\int \cos 2 x d x \\ & =-\frac{1}{2} \sin 2 x+C .\end{aligned}$
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