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

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Aicharya · Accepted Answer

$\begin{aligned}
& \int x^3 \sin x d x=-x^3 \cos x+\int 3 x^2 \cos x d x \\ & =-x^3 \cos x+3 x^2 \sin x-\int 6 x \sin x d x \\ & =-x^3 \cos x+3 x^2 \sin x+6 x \cos x-6 \sin x+c
\end{aligned}$
So $g(x)=-x^3 \cos x+3 x^2 \sin x+6 x \cos x-6 \sin x$
$\begin{aligned}
& \mathrm{g}\left(\frac{\pi}{2}\right)=\frac{3 \pi^2}{4}-6 \\ & \mathrm{~g}^{\prime}(\mathrm{x})=-3 \mathrm{x}^2 \cos \mathrm{x}+\mathrm{x}^3 \sin \mathrm{x}+6 \cos \mathrm{x}-6 \cos \mathrm{x} \\ & \mathrm{~g}^{\prime}\left(\frac{\pi}{2}\right)=\frac{\pi^3}{8} \\ & 8\left(\mathrm{~g}\left(\frac{\pi}{2}\right)+\mathrm{g}^{\prime}\left(\frac{\pi}{2}\right)\right)=\pi^3+6 \pi^2-48
\end{aligned}$
So $\alpha+\beta-\gamma=55$

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

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