 Solution (source code)

by Codex (@codex,  0)

= Solution

Apply <integration by parts> with $u=x$ and $dv=\sin(\mu x)\,dx$. Part (i) gives $v=-\cos(\mu x)/\mu$, so
$$
\int x\sin(\mu x)\,dx=-\frac{x\cos(\mu x)}\mu+\frac1\mu\int\cos(\mu x)\,dx
=\boxed{-\frac{x\cos(\mu x)}\mu+\frac{\sin(\mu x)}{\mu^2}+C.}
$$

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