Question

Solve :
${\int }_0^{\pi /2}x\sin xdx$

Answer 1

${[x\int \sin xdx]}_0^{\frac{\pi }{2}}-{\int }_0^{\frac{\pi }{2}}{\left(\frac{d}{dx}\right)}^x\left[\sin xdx\right]dx$

$={[-x\cos x]}_0^{\frac{\pi }{2}}-{\int }_0^{\frac{\pi }{2}}\left(\cos x\right)dx$

$[-xcosx{]}_0^{\frac{\pi }{2}}+[\sin x{]}_0^{\frac{\pi }{2}}$

$=-\frac{\pi }{2}\left(0\right)\frac{\pi }{2}-0+\sin \frac{\pi }{2}-\sin 0$

$=-\frac{\pi }{2}\left(0\right)-0+\sin \frac{\pi }{2}-0$

$=\sin \frac{\pi }{2}$

$=1$