Let x^3 xdx=g(x)+C, where C is the constant of integration.

Question

Let $\int {x}^{3}\sin xdx=g(x)+C$, where $C$ is the constant of integration. If $8(g(\frac{\pi }{2})+{g}^{'}(\frac{\pi }{2}))=\alpha {\pi }^{3}+\beta {\pi }^{2}+\gamma ,\alpha ,\beta ,\gamma \in Z$, then $\alpha +\beta -\gamma $ equals :

Accepted Answer

$\begin{gathered} \int {x}^{3}\sin xdx=-{x}^{3}\cos x+\int 3{x}^{2}\cos xdx \ =-{x}^{3}\cos x+3{x}^{2}\sin x-\int 6x\sin xdx \ =-{x}^{3}\cos x+3{x}^{2}\sin x+6x\cos x-6\sin x+c \end{gathered}$

So $g(x)=-{x}^{3}\cos x+3{x}^{2}\sin x+6x\cos x-6\sin x$

$\begin{gathered} g(\frac{\pi }{2})=\frac{3{\pi }^{2}}{4}-6 \ {g}^{'}(x)={x}^{3}\sin x \ {g}^{'}(\frac{\pi }{2})=\frac{{\pi }^{3}}{8} \ 8(g(\frac{\pi }{2})+{g}^{'}(\frac{\pi }{2}))={\pi }^{3}+6{\pi }^{2}-48 \end{gathered}$

So $\alpha +\beta -\gamma =55$

Solutions · 1