How can I solve this integral by making any substitution?

Question

The integral is $$\int{\left[\frac{\sin^8(x) - \cos^8(x)}{1 - 2 \sin^2(x)\cos^2(x)}\right]}dx$$

Answer 1

Hint: $\sin^8(x)- \cos^8(x)=(\sin^2(x)- \cos^2(x))(1-2\sin^2(x) \cos^2(x))$

Answer 2

this integer can be reduced using the double angle sine and cosine formula.

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