question_answer

Evaluate \[\int_{0}^{\pi /2}{x\,\sin \,x\,dx.}\]

Answer:

Let \[l=\int_{0}^{\pi /2}{\underset{I}{\mathop{x}}\,}\underset{II}{\mathop{\sin }}\,xdx.\] Then, \[=[-\,x\cos \,x]_{0}^{\pi /2}-\int_{0}^{\pi /2}{1\times (-\cos \,x)\,dx}\]                  [Integrating by parts] \[=[-\,x\cos \,x]_{0}^{\pi /2}+[\sin ]_{0}^{\pi /2}\] \[=\left( -\frac{\pi }{2}\cos \frac{\pi }{2}+0\cos 0 \right)+\left( \sin \frac{\pi }{2}-\sin 0 \right)=1\]