Question

Evaluate :

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

Answer 1

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

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

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

$I=-\int \cos 2xdx=\frac{-1}{2}\sin 2x+c$