Answer
\[ = \pi \]
Work Step by Step
\[\begin{gathered}
\int_0^\pi {x\sin x\,dx} \hfill \\
\hfill \\
set\,\,\,the\,\,substitution \hfill \\
\hfill \\
u = x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,du = dx \hfill \\
dv = \sin xdx\,\,\,\, \to \,\,\,v = - \cos x \hfill \\
\hfill \\
use\,\,uv - \int_{}^{} {vdu} \hfill \\
\hfill \\
{\text{replacing the values }}{\text{in the equation}} \hfill \\
\hfill \\
= - x\cos x - \int_{}^{} {\,\left( { - \cos x} \right)dx} \hfill \\
\hfill \\
integrate\,\, \hfill \\
\hfill \\
- x\cos x + \sin x + C \hfill \\
\hfill \\
{\text{Therefore}}{\text{,}} \hfill \\
\int_0^\pi {x\sin x\,dx} = \,\,\left[ { - x\cos x + \sin x} \right]_0^\pi \hfill \\
\hfill \\
evaluate\,\,the\,\,{\text{limits}} \hfill \\
\hfill \\
= \,\left( { - \pi \cos \pi + \sin \pi } \right) - \,\left( 0 \right) = \pi \hfill \\
\end{gathered} \]
