Answer
$\int x\cos 4x\,dx=\frac{1}{4}x\sin(4x)+\frac{1}{16}\cos(4x)+C$
Work Step by Step
$$A=\int x\cos 4xdx$$
We would choose $u=x$ and $dv=\cos 4xdx$
We could easily see that $du=dx$.
For $dv=\cos 4xdx$, we analyze as follows $$\int \cos 4xdx$$
Let $u=4x$, then $du=4dx$, so $dx=\frac{1}{4}du$ $$\int\cos 4xdx=\frac{1}{4}\int\cos udu=\frac{1}{4}\sin u+C=\frac{1}{4}\sin(4x)+C$$
Therefore, for $dv=\cos 4xdx$, $v=\frac{1}{4}\sin(4x)$
Apply Integration by Parts to A, we have $$A=uv-\int vdu$$ $$A=\frac{1}{4}x\sin(4x)-\int\frac{1}{4}\sin(4x)dx$$ $$A=\frac{1}{4}x\sin(4x)-\frac{1}{4}\int\frac{1}{4}\sin(4x)d(4x)$$ $$A=\frac{1}{4}x\sin(4x)-\frac{1}{16}(-\cos(4x)+C)$$ $$A=\frac{1}{4}x\sin(4x)+\frac{1}{16}\cos(4x)+C$$
