Решить неопределенный интеграл (sin2x)^4dx

Решите задачу:

$\int\ {(sin2x)^4} \, dx \\ 2x=m \\ dx= \frac{1}{2}dm \\ = \frac{1}{2} \int\sin^{4}m \, dm= \frac{1}{2} \int\ \frac{(1-cos2m)^2}{2^2} } \, dm \\ = \frac{1}{8} \int\((1-2cos2m+cos^{2}2m) \, dm= \\ = \frac{1}{8}m- \frac{1}{4}* \frac{1}{2}sin2m+ \frac{1}{8} \int\ {cos^{2}2m} \, dm \\ = \frac{1}{8}m- \frac{1}{8}sin2m+ \frac{1}{8} \int\ { \frac{1+cos4m}{2} } \, dm \\ = \frac{1}{8}m- \frac{1}{8}sin2m+ \frac{1}{16}m+ \frac{1}{16}* \frac{1}{4}sin4m \\$
$= \frac{1}{8}2x- \frac{1}{8}sin4x+ \frac{1}{16}2x+ \frac{1}{16} \frac{1}{4}sin8x \\ = \frac{x}{4}- \frac{sin4x}{8}+ \frac{x}{8}+ \frac{sin8x}{64} \\ = \frac{3x}{8} - \frac{sin4x}{8}+ \frac{sin8x}{64}+C$