Explanation

(a)
$\int_{}^{}{\frac{\sin^{8}{x - \cos^{8}x}}{1 - 2\sin^{2}{x\ \cos^{2}x}}\ dx}$
$= \int_{}^{}{\frac{{(\sin^{4}}{x - \cos^{4}{x)\left( \sin^{4}{x + \cos^{4}x} \right)}}}{(1 - 2\sin^{2}{x\ \cos^{2}{x)}}}\ dx}$
$= \int_{}^{}{\frac{{(\sin^{4}}{x - \cos^{4}{x)\left( \sin^{4}{x + \cos^{4}x} \right)}}}{(1 - 2\sin^{2}{x\ \cos^{2}{x)}}}\ \times \lbrack\left( \sin^{2}{x\ \cos^{2}x} \right)^{2}} - 2\sin^{2}{x\ \cos^{2}{x\rbrack}}dx\$
$= - \int_{}^{}\frac{\cos{2x(1 - 2\sin^{2}{x\cos^{2}{x)}}}}{(1 - 2\sin^{2}{x\cos^{2}{x)}}}\ dx$
$= - \int_{}^{}{\cos{2x}}\ dx\$
$\Rightarrow A\sin{2x + B = - \frac{1}{2}\sin{2x + B}}$ (given)
$\therefore\ A = - \frac{1}{2}$