Let \(\int \mathrm{x}^3 \sin \mathrm{xdx}=\mathrm{g}(\mathrm{x})+\mathrm{C}\), where C is the constant of integration. I…

\(\text{Using Integration by parts}\\ \int {x}^{3}\sin xdx=-{x}^{3}\cos x+\int 3{x}^{2}\cos xdx\\ =-{x}^{3}\cos x+3{x}^{2}\sin x-\int 6x\sin xdx\\ \text{ So }g(x)=-{x}^{3}\cos x+3{x}^{2}\sin x\\ +6x\cos x-6\sin x\\ g\left(\frac{\pi }{2}\right)=\frac{3{\pi }^{2}}{4}-6\\ {g}^{'}(x)=-3{x}^{2}\cos x+{x}^{3}\sin x\\ +3{x}^{2}\cos x+6\cos x-6\cos x\\ {g}^{'}\left(\frac{\pi }{2}\right)=\frac{{\pi }^{3}}{8}\\ 8\left(g\left(\frac{\pi }{2}\right)+{g}^{'}\left(\frac{\pi }{2}\right)\right)={\pi }^{3}+6{\pi }^{2}-48\\ \text{ So }\alpha +\beta -\gamma =55\)