Let $\int x^3\sin x\,dx=g(x)+C$, where $C$ is the constant of integration. If $8\left(g\left(\frac\pi2\right)+g'\left(\frac\pi2\right)\right)=\alpha\pi^3+\beta\pi^2+\gamma$, $\alpha,\beta,\gamma\in Z$, then $\alpha+\beta-\gamma$ equals :

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] $g(x)=-x^3\cos x+3x^2\sin x+6x\cos x-6\sin x$, so $g\left(\frac\pi2\right)=\frac{3\pi^2}4-6$, and $g'(x)=x^3\sin x$ gives $g'\left(\frac\pi2\right)=\frac{\pi^3}8$. $8(g+g')=\pi^3+6\pi^2-48$, so $\alpha=1,\beta=6,\gamma=-48$ and $\alpha+\beta-\gamma=55$. Answer (2).

Q. Let $\int x^3\sin x\,dx=g(x)+C$, where $C$ is the constant of integration. If $8\left(g\left(\frac\pi2\right)+g'\left(\frac\pi2\right)\right)=\alpha\pi^3+\beta\pi^2+\gamma$, $\alpha,\beta,\gamma\in Z$, then $\alpha+\beta-\gamma$ equals :

Explanation

Reviewed by Dr. Abubacker

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