$\int \frac{{\sin }^8\mathrm{x}-{\cos }^8\mathrm{x}}{1-2{\sin }^2{\mathrm{xcos}}^2\mathrm{x}}\mathrm{dx}=$

$\begin{array}{l}I=\int \frac{{\sin }^{8}x-{\cos }^{8}x}{1-2{\sin }^{2}x{\cos }^{2}x}dx\\ =\int \frac{\left({\sin }^{4}x-{\cos }^{4}x\right)\left({\sin }^{4}x+{\cos }^{4}x\right)}{1-2{\sin }^{2}x{\cos }^{2}x}dx\end{array}$
$\begin{array}{c}=\int \frac{\left({\sin }^{4}x+{\cos }^{4}x\right)\left({\sin }^{2}x-{\cos }^{2}x\right)\left({\sin }^{2}x+{\cos }^{2}x\right)}{1-2{\sin }^{2}x{\cos }^{2}x}dx\\ =\int \frac{1\left({\sin }^{2}x-{\cos }^{2}x\right)\left[{\left({\sin }^{2}x+{\cos }^{2}x\right)}^2\right.{\displaystyle \left.-2{\sin }^{2}x{\cos }^{2}x\right]}}{1-2{\sin }^{2}x{\cos }^{2}x}dx\\ =\int \frac{\left({\sin }^{2}x-{\cos }^{2}x\right)\left(1-2{\sin }^{2}x{\cos }^{2}x\right)}{1-2{\sin }^{2}x{\cos }^{2}x}dx\\ =-\int \cos 2xdx=-\frac{1}{2}\sin 2x+c\end{array}$