Thomas' Calculus 13th Edition

$-2x\cos(\frac{x}{2})+4\sin(\frac{x}{2})+C$
We use the formula $\int udv=uv-\int vdu$ with $u=x,\,dv=\sin \frac{x}{2}dx$ $du=dx,\,v=-2\cos(\frac{x}{2})$ Then, $\int x\sin\frac{x}{2}dx=x\times-2\cos(\frac{x}{2})-\int-2\cos(\frac{x}{2})dx$ $=-2x\cos(\frac{x}{2})+4\sin(\frac{x}{2})+C$