karush said:
$\tiny{9.1}$
\begin{align*} \displaystyle
I&=\int -3x \cos{4x} \, dx = -3\int x \cos{4x} \, dx\\
\textit{uv substitution}\\
u&=x\therefore du=dx\\
dv&=\cos{4x} \, dx\therefore v=-\frac{\sin{4x}}{4}\\
\textit{now we have}\\
I&=3\left[\frac{x\sin{4x}}{4}
+\int\frac{\sin{4x}}{4} \, dx\right]\\
\textit{and finally}\\
I&=\frac{3}{4} x\sin{4x} +\frac{1}{16}\cos{4x}+C
\end{align*}
think this is ok but probably suggestions
Another option - start by differentiating your function twice
$\displaystyle \begin{align*} \frac{\mathrm{d}^2}{\mathrm{d}x^2}\,\left[ -3\,x\cos{ \left( 4\,x \right) } \right] &= \frac{\mathrm{d}}{\mathrm{d}x} \,\left[ 12\,x\sin{ \left( 4\,x \right) } - 3 \cos{ \left( 4\,x \right) } \right] \\ &= 48\,x\cos{ \left( 4\,x \right) } + 12\sin{ \left( 4\,x \right) } + 12 \cos{ \left( 4\,x \right) } \end{align*}$
and since every derivative equation can be written in an equivalent form as an integral, as $\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \,\left[ 12\,x\sin{ \left( 4\,x \right) } - 3\cos{ \left( 4\,x \right) } \right] &= 48\,x\cos{ \left( 4\,x \right) } + 12\sin{ \left( 4\,x \right) } + 12\cos{ \left( 4\,x \right) } \end{align*}$ that means
$\displaystyle \begin{align*} \int{ \left[ 48\,x\cos{ \left( 4\,x \right) } + 12\sin{ \left( 4\,x \right) } + 12\cos{ \left( 4\,x \right) } \right] \,\mathrm{d}x } &= 12\,x\sin{ \left( 4\,x \right) } - 3\cos{ \left( 4\,x \right) } + C_1 \\ \int{ 48\cos{ \left( 4\,x \right) } \,\mathrm{d}x} + \int{12\sin{ \left( 4\,x \right) } \,\mathrm{d}x} + \int{ 12\cos{ \left( 4\,x \right) } \,\mathrm{d}x} &= 12\,x\sin{ \left( 4\,x \right) } - 3\cos{ \left( 4\,x \right) } + C_1 \\ \int{48\,x\cos{ \left( 4\,x \right) } \,\mathrm{d}x} - 3\cos{ \left( 4\,x \right) } + C_2 + 3\sin{ \left( 4\,x \right) } + C_3 &= 12\,x \sin{ \left( 4\,x \right) } - 3\cos{ \left( 4\,x \right) } + C_1 \\ \int{ 48\,x\cos{ \left( 4\,x \right) }\,\mathrm{d}x} &= 12\,x\sin{ \left( 4\,x \right) } - 3\cos{ \left( 4\,x \right) } + 3\cos{ \left( 4\,x \right) } - 3\sin{ \left( 4\,x \right) } + C_1 - C_2 - C_3 \\ -16 \int{ -3\,x\cos{ \left( 4\,x \right) } \,\mathrm{d}x } &= 12\,x\sin{ \left( 4\,x \right) } - 3\sin{ \left( 4\,x \right) } + C_1 - C_2 - C_3 \\ \int{ -3\,x\cos{ \left( 4\,x \right) } \,\mathrm{d}x} &= \frac{3}{16}\,\sin{ \left( 4\,x \right) } - \frac{3}{4}\,x\sin{ \left( 4\,x \right) } + C \textrm{ where } C = -\frac{1}{16}\,\left( C_1 - C_2 - C_3 \right) \end{align*}$