$\int (x+3)\sin\dfrac{x}{2}\cos\dfrac{x}{2} \, dx\\ =\dfrac{1}{2}\int (x+3)(\sin0 + \sin x) \, dx\\ =\dfrac{1}{2}\int (x+3)\sin x \, dx\\ u=x+3=>du=dx\\ dv=\sin x dx=>v=-\cos x\\ \dfrac{1}{2}\int (x+3)\sin x \, dx=\dfrac{1}{2}(-(x+3)\cos x + \int\cos x \, dx)\\=\dfrac{-1}{2}(x+3)\cos x +\sin x +C$