^8x-^8x1-2^2x^2x,dx is equal to:

Question

$\displaystyle\int \dfrac{\sin^8x-\cos^8x}{1-2\sin^2x\cos^2x}\,dx$ is equal to:

Accepted Answer

$I = \int \dfrac{(\sin^4x)^2-(\cos^4x)^2}{1-2\sin^2x\cos^2x}\,dx = \int \dfrac{(\sin^4x+\cos^4x)(\sin^4x-\cos^4x)}{1-2\sin^2x\cos^2x}\,dx$

$= \int \dfrac{[(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x](\sin^2x+\cos^2x)(\sin^2x-\cos^2x)}{1-2\sin^2x\cos^2x}\,dx$

$= -\int \cos 2x\,dx = -\dfrac{\sin 2x}{2}+c$

Solutions · 1