x^3 x,dx = g(x)+C.

Question

$\displaystyle\int x^3\sin x\,dx = g(x)+C$. If $8\left(g\left(\dfrac\pi2\right)+g'\left(\dfrac\pi2\right)\right) = \alpha\pi^3+\beta\pi^2+\gamma$, $\alpha,\beta,\gamma\in\mathbb{Z}$, then $\alpha+\beta-\gamma$ is equal to:

Accepted Answer

By repeated integration by parts:

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

$$g\left(\frac\pi2\right) = \frac{3\pi^2}4-6,\qquad g'(x)=x^3\sin x \implies g'\left(\frac\pi2\right)=\frac{\pi^3}8$$

$$8\left(g\left(\frac\pi2\right)+g'\left(\frac\pi2\right)\right) = \pi^3+6\pi^2-48$$

So $\alpha=1,\,\beta=6,\,\gamma=-48$, giving $\alpha+\beta-\gamma = 1+6+48=55$.

Solutions · 1