Answer
$$\int \sin ^{4} 6 \theta d \theta= \frac{1}{4} \left (\frac{3}{2}\theta-\frac{1}{6}\sin12\theta +\frac{1}{48}\sin24\theta\right)+c$$
Work Step by Step
$$
\int \sin ^{4} 6 \theta d \theta
$$
Since \begin{align*}
\sin^2x&=\frac{1}{2}(1-\cos2x)\\
\cos^2x&=\frac{1}{2}(1+\cos2x)
\end{align*}
Then
\begin{align*}
\int \sin ^{4} 6 \theta d \theta&= \frac{1}{4}\int (1-\cos12\theta)^2d\theta\\
&= \frac{1}{4}\int (1-2\cos12\theta+\cos^212\theta)d\theta\\
&= \frac{1}{4}\int\left (1-2\cos12\theta+\frac{1}{2}+\frac{1}{2}\cos24\theta\right)d\theta\\
&= \frac{1}{4}\int\left (\frac{3}{2}-2\cos12\theta +\frac{1}{2}\cos24\theta\right)d\theta\\
&= \frac{1}{4} \left (\frac{3}{2}\theta-\frac{1}{6}\sin12\theta +\frac{1}{48}\sin24\theta\right)+c
\end{align*}
