JEE Main 2025 — Indefinite Integration Question with Solution

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