I know an answer's already been accepted, but I figured I'd chime in with an answer that doesn't require nice bounds and isn't quite as ugly as Ragib's dark path. To get a more workable denominator, multiply numerator and denominator by $\cos x-\sin x$.
$\int \frac{\sin^2x\cos^2x-\sin^3x\cos x}{\cos^2x-sin^2x}dx=\int\frac{\frac14\sin^22x-\frac12\sin2x(1-\cos^2x)}{\cos2x}dx=$
$\int \frac{\frac14-\frac14\cos^22x-\frac12\sin2x(1-\frac{1+\cos2x}2)}{\cos2x}dx=\int\frac{\frac14-\frac14\cos^22x-\frac14\sin2x+\frac14\sin2x\cos2x}{\cos2x}dx=$
$\int \frac14\sec2x-\frac14\cos2x+\frac18(\frac{-2\sin2x}{\cos2x})+\frac14\sin2xdx=$
$\frac18\ln|\sec2x+\tan2x|-\frac18\sin2x+\frac18\ln|\cos2x|-\frac18\cos2x+C=$
$\frac18\ln|1+\sin2x|-\frac18\sin2x-\frac18\cos2x+C$
If you plug in the bounds given for the problem, you'll find all the sine terms equal $0$, leaving $-\frac18(\cos\pi-\cos0)=-\frac18(-1-1)=\frac14$, matching Ragib's answer.