Answer
$$ -\frac{3}{4} x^{2} \sin 8 x-\frac{3}{16} x \cos 8 x+\frac{3}{128} \sin 8 x+C$$
Work Step by Step
Given $$\int-6 x^{2} \cos 8 x d x$$
Use integration by parts
\begin{align*}
u&= -6x^2\ \ \ \ \ \ \ dv= \cos 8 x d x\\
du&= -12xdx \ \ \ \ \ \ \ v= \frac{1}{8}\sin 8x
\end{align*}
Then
\begin{align*}
\int-6 x^{2} \cos 8 x d x&= \frac{-6x^2}{8}\sin 8x +\frac{12}{8}\int x\sin 8xdx\\
\end{align*}
To integrate $ \displaystyle\int x\sin 8xdx$ Use integration by parts again
\begin{align*}
u&= x\ \ \ \ \ \ \ dv= \sin 8 x d x\\
du&= dx \ \ \ \ \ \ \ v= \frac{-1}{8}\cos 8x
\end{align*}
Then
\begin{align*}
\int x\sin 8xdx&= x\left(-\frac{1}{8} \cos 8 x\right)-\int\left(-\frac{1}{8} \cos 8 x\right) d x\\
&=\frac{-1}{8}x\cos 8x +\frac{1}{8} \sin 8x+c
\end{align*}
Hence
$$\int-6 x^{2} \cos 8 x d x=-\frac{3}{4} x^{2} \sin 8 x-\frac{3}{16} x \cos 8 x+\frac{3}{128} \sin 8 x+C$$
