Answer
$$\int \csc ^{4} 3 x d x =\frac{-1}{3}\cot3x -\frac{1}{9}\cot^33x$$
Work Step by Step
$$
\int \csc ^{4} 3 x d x
$$
Let $u=3x\ \ \to\ \ du=3dx$, then
\begin{align*}
\int \csc ^{4} 3 x d x&=\frac{1}{3}\int \csc ^{4} u d u\\
&=\frac{1}{3}\int \csc ^{2} u \csc ^{2} ud u\\
&=\frac{1}{3}\int(1+ \cot ^{2} u) \csc ^{2} ud u\\
&=\frac{1}{3}\int(\csc ^{2} u+ \cot ^{2} u\csc ^{2} u) d u\\
&=\frac{1}{3}\left( -\cot u -\frac{1}{3}\cot^3u\right)+c\\
&=\frac{-1}{3}\cot3x -\frac{1}{9}\cot^33x
\end{align*}
