- Online homework says this is wrong. ??
NO! $\displaystyle \begin{align*} 9x\cdot \frac{1}{8}\sin{(8x)} \end{align*}$ is NOT $\displaystyle \begin{align*} \frac{9}{8}\sin^2{(8x)} \end{align*}$, it's $\displaystyle \begin{align*} \frac{9}{8}x\sin{(8x)} \end{align*}$.
Also $\displaystyle \begin{align*} \int{\frac{9}{8}\sin{(8x)}\,\mathrm{d}x} = -\frac{1}{8}\cdot \frac{9}{8}\cos{(8x)} = -\frac{9}{64}\cos{(8x)} \end{align*}$, NOT $\displaystyle \begin{align*} \frac{72}{8}\cos{(8x)} \end{align*}$.