Question

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

Answer 1

$\frac{sin^8x-cos^8x}{1-2sin^2xcos^2x}$=$\frac{(sin^4x+cos^4x)(sin^4x-cos^4x)}{sin^2x+cos^2x-sin^2xcos^2x-sin^2xcos^2x}$

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

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

=-cos2x

∴$\frac{sin^8x-cos^8x}{1-2sin^2xcos^2x}$=∫-cos2xdx=-sin2$\frac{x}{2}$+C