\int \sin^2 ax \cos^2 bx dx = \frac{x}{4}
-\frac{\sin 2ax}{8a}-
\frac{\sin[2(a-b)x]}{16(a-b)}
+\frac{\sin 2bx}{8b}-
\frac{\sin[2(a+b)x]}{16(a+b)}+C