If ${\displaystyle {\int }_0^{\frac{\pi }{2}}{x}^n}\sin xdx=\left(\frac{3}{4}\right)\left({\pi }^2-8\right),$  then the value of n is ..........

${\displaystyle {\int }_0^{\frac{\pi }{2}}{x}^3}\sin xdx$

$={\left[x^3\left(-\cos x\right)\right]}_0^{\frac{\pi }{2}}-{\displaystyle {\int }_0^{\frac{\pi }{2}}3x^2\left(-\cos x\right)dx}$

$=3{\displaystyle {\int }_0^{\frac{\pi }{2}}{x}^2\cos xdx}$

$=3{\left[x^2\left(\sin x\right)\right]}_0^{\frac{\pi }{2}}-{\displaystyle {\int }_0^{\frac{\pi }{2}}2x\sin xdx}$

$=3\left[\frac{{\pi }^2}{4}-2{\displaystyle {\int }_0^{\frac{\pi }{2}}x\sin xdx}\right]$

$=3\frac{{\pi }^2}{4}-6\left[x{\left(-\cos x\right)}_0^{\frac{\pi }{2}}-{\displaystyle {\int }_0^{\frac{\pi }{2}}\left(-\cos x\right)dx}\right]$

$=\frac{3{\pi }^2}{4}-6{\displaystyle {\int }_0^{\frac{\pi }{2}}\left(+\cos x\right)dx}$

$=\frac{3{\pi }^2}{4}-6{\left(\sin x\right)}_0^{\frac{\pi }{2}}$

$=\frac{3{\pi }^2}{4}-6=3\left(\frac{{\pi }^2}{4}-2\right)$

$=3\left(\frac{{\pi }^2-8}{4}\right)=\frac{3}{4}\left({\pi }^2-8\right)$

$\boxed{\therefore n=3}$ .