\begin{gather*} \int x\cdot\sin x\;\mathrm{d}x=\left|\begin{array}{cc} u=x & v=-\cos x\\ u'=1 & v=\sin x\end{array}\right|=-x\cdot\cos x-\int-\cos x\;\mathrm{d}x=\\ =-x\cdot\cos x+\sin x\end{gather*}