∫ sin 8 x - cos 8 x 1 - 2 sin 2 x cos 2 x d x is equal to

${\sin }^{8}x-{\cos }^{8}x=\left({\sin }^{2}x-{\cos }^{2}x\right)\left({\sin }^{2}x+{\cos }^{2}x\right)\left({\sin }^{4}x+{\cos }^{4}x\right)$

and ${\sin }^{4}x+{\cos }^{4}x={\left({\sin }^{2}x+{\cos }^{2}x\right)}^2-2{\sin }^{2}x{\cos }^{2}x=1-2{\sin }^{2}x{\cos }^{2}x$

$\therefore I=\int \frac{{\sin }^{8}x-{\cos }^{8}x}{\left(1-2{\sin }^{2}x{\cos }^{2}x\right)}dx$

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

 $=-\int \cos 2xdx$ 

  $=-\frac{1}{2}\sin 2x+c$, where $c$ is the constant of integration.