Question

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

Answer 1

${\int }_0^{\frac{\pi }{2}}{x}^2\sin xdx$

Let $u=x^2\Rightarrow du=2xdx$

$dv=\sin xdx\Rightarrow v=-\cos x$

$={[-x^2\cos x]}_0^{\frac{\pi }{2}}+2{\int }_0^{\frac{\pi }{2}}x\cos xdx$

Let $u=x\Rightarrow du=dx$

$dv=\cos xdx\Rightarrow v=\sin x$

$=-[{\left(\frac{\pi }{2}\right)}^2\cos \frac{\pi }{2}-0]+2{\left[x\sin x\right]}_0^{\frac{\pi }{2}}-2{\int }_0^{\frac{\pi }{2}}\sin xdx$

$=0+2[\frac{\pi }{2}\sin \frac{\pi }{2}-0]-2{[-\cos x]}_0^{\frac{\pi }{2}}$

$=\pi +2[\cos \frac{\pi }{2}-\cos 0]$

$=\pi +2[0-1]=\pi -2$