Answer
$$-\frac{1}{12} \sin ^{3}(3 x) \cos (3 x)-\frac{1}{8} \sin (3 x) \cos (3 x)+\frac{3}{8} x+c$$
Work Step by Step
Given $$\int \sin ^{4}(3 x) d x$$
Let $$u=3x\ \ \Rightarrow \ du= 3dx $$
Then by using
$$\int \sin ^{n} x d x=-\frac{\sin ^{n-1} x \cos x}{n}+\frac{n-1}{n} \int \sin ^{n-2} x d x$$
\begin{align*}
\int \sin ^{4}(3 x) d x&=\frac{1}{3} \int \sin ^{4} u d u\\
&=-\frac{1}{12} \sin ^{3} u \cos u+\frac{1}{4} \int \sin ^{2} u d u \quad(n=4)\\
&=-\frac{1}{12} \sin ^{3} u \cos u+\frac{1}{4}\left(-\frac{1}{2} \sin u \cos u+\frac{1}{2} \int d u\right) \quad(n=2)\\
&=-\frac{1}{12} \sin ^{3} u \cos u-\frac{1}{8} \sin u \cos u+\frac{1}{8} u+c\\
&=-\frac{1}{12} \sin ^{3}(3 x) \cos (3 x)-\frac{1}{8} \sin (3 x) \cos (3 x)+\frac{3}{8} x+c
\end{align*}
