First, the substitution \(w = 3x\) will simplify the integral to
\begin{equation*}
\int \sin^2(3x)\cos^2(3x) dx = \frac{1}{3}\int\sin^2(w)\cos^2(w) dw.
\end{equation*}
\begin{align*}
\int \sin^2(w)\cos^2(w)dw &=
\int\frac{1-\cos(2w)}{2}\frac{1+\cos(2w)}{2} dw \\
& = \frac{1}{4} \int 1-\cos^2(2w)
\end{align*}
Apply
(A.14) to
\(\cos^2(2w)\) and we get
\begin{align*}
\frac{1}{4} \int 1-\cos^2(2w) dw & = \frac{1}{4} \int
1- \frac{1+\cos(4w)}{2} dw = \frac{1}{4}\int \frac{1}{2} -
\frac{\cos(4w)}{2} dw \\
&= \frac{1}{8}\int 1 - \cos(4w) dw\\
& = \frac{1}{8} \left(w - \frac{1}{4}\sin(4w)\right) +C\\
& = \frac{3x}{8} - \frac{\sin(12x)}{32} + C
\end{align*}
Putting all these together, we get
\begin{align*}
\int \sin^2(3x)\cos^2(3x) dx & =
\frac{1}{3}\int\sin^2(w)\cos^2(w) dw.\\
& = \frac{1}{3}\left( \frac{3x}{8} - \frac{\sin(12x)}{32}\right)+C\\
& = \frac{x}{8} - \frac{\sin(12x)}{96} + C
\end{align*}