Hello,
\(\boxed{sin^4\ x= \frac{3}{8}- \frac{cos\ 2x}{2}+\frac{sin\ 4x}{8} }\\
Let's assume t=6x \ dt=6\ dx\\
\int\limits {sin^4\ 6x} \, dx = \frac{1}{6} *\int\limits {sin^4\ t} \, dt \\
= \dfrac{1}{6} *( \frac{3t}{8} - \frac{sin\ 2t}{4} +\frac{sin\ 4t}{32})+C\\
= \dfrac{3x}{8} - \frac{sin\ 12x}{24} +\frac{sin\ 24x}{192} +C\)
