Question

Integrate the function $\frac{{\sin }^{8}x-{\cos }^{8}x}{1-2{\sin }^{2}x{\cos }^{2}x}$

Answer 1

$\frac{{\sin }^{8}x-{\cos }^{8}x}{1-2{\sin }^{2}x{\cos }^{2}x}=\frac{({\sin }^{4}x+{\cos }^{4}x)({\sin }^{4}x-{\cos }^{4}x)}{{\sin }^{2}x+{\cos }^{2}x-{\sin }^{2}x{\cos }^{2}x-{\sin }^{2}x{\cos }^{2}x}$
$=\frac{({\sin }^{4}x+{\cos }^{4}x)({\sin }^{2}x+{\cos }^{2}x)({\sin }^{2}x-{\cos }^{2}x)}{({\sin }^{2}x-{\sin }^{2}x{\cos }^{2}x)+({\cos }^{2}x-{\sin }^{2}x{\cos }^{2}x)}$
$=\frac{({\sin }^{4}x+{\cos }^{4}x)({\sin }^{4}x-{\cos }^{2}x)}{{\sin }^{2}x(1-{\cos }^{2}x)+{\cos }^{2}x(1-{\sin }^{2}x)}$
$=\frac{-({\sin }^{4}x+{\cos }^{4}x)({\cos }^{2}x-{\sin }^{2}x)}{({\sin }^{4}x+{\cos }^{4}x)}=-\cos 2x$
$\therefore \int \frac{{\sin }^{8}x-{\cos }^{8}x}{1-2{\sin }^{2}x{\cos }^{2}x}dx=\int -\cos 2xdx=-\frac{\sin 2x}{2}+C$