$\textstyle \int \sin^42x\mathrm dx$$=\int \left ( \frac{1-\cos 4x}{2} \right )^2\mathrm dx$
$=\frac 14\int\left ( \cos^2 4x-2\cos 4x +1\right )\mathrm dx$
$=\frac 18\int\left (1+\cos 8x-4\cos 4x +2\right )\mathrm dx$
$=\frac{\sin 8x}{64}-\frac{\sin4x}{8}+\frac{3x}{8}+C$