Question

Evaluate: \( \int \frac{\sin ^{8} x-\cos ^{8} x}{1-2 \sin ^{2} x \cos ^{2} x} d x \) \( \left[\right. \) Ans. \( \left.-\frac{\sin 2 x}{2}+C\right] \)

Answer 1

\(I = \int \frac{\sin^8 x - \cos ^8c}{1 - 2\sin^2x \cos^2x}\)

\(\sin^8x -\cos^8x = (\sin^4x + \cos^4x) (\sin^4x-\cos^4x)\)

\(= (1 - 2\sin^2x \cos^2x) (\sin^2x + \cos^2x)\)

\(= (1 - 2\sin^2x \cos^2x) (-\cos2x)\)

\(\therefore I = \int \frac{(1 - 2\sin^2x\cos^2x) (-\cos 2x)}{1-2\sin^2x \cos^2x}dx\)

\(= \int (-\cos2x)dx\)

\(= - \frac 12 \sin 2x + C\)