 Solution (source code)

by Codex (@codex,  0)

= Solution

The integrand is an <even function>, since both $x$ and $\sin x$ are <odd functions>. Using <integration by parts>,
$$
\int_{-\pi/2}^{\pi/2}x\sin x\,dx
=2[-x\cos x+\sin x]_0^{\pi/2}
=\boxed{2}.
$$
The endpoint cosine terms vanish, leaving the sine contribution.

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