$\begin{aligned}
& \int x^3 \sin x d x=-x^3 \cos x+\int 3 x^2 \cos x d x \
& =-x^3 \cos x+3 x^2 \sin x-\int 6 x \sin x d x \
& =-x^3 \cos x+3 x^2 \sin x+6 x \cos x-6 \sin x+c
\end{aligned}Sog(x)=-x^3 \cos x+3 x^2 \sin x+6 x \cos x-6 \sin x\begin{aligned}
& \mathrm{g}\left(\frac{\pi}{2}\right)=\frac{3 \pi^2}{4}-6 \
& \mathrm{g}^{\prime}(\mathrm{x})=-3 \mathrm{x}^2 \cos \mathrm{x}+\mathrm{x}^3 \sin \mathrm{x}+6 \cos \mathrm{x}-6 \cos \mathrm{x} \
& \mathrm{g}^{\prime}\left(\frac{\pi}{2}\right)=\frac{\pi^3}{8} \
& 8\left(\mathrm{~g}\left(\frac{\pi}{2}\right)+\mathrm{g}^{\prime}\left(\frac{\pi}{2}\right)\right)=\pi^3+6 \pi^2-48
\end{aligned}So\alpha+\beta-\gamma=55$
Let ∫ x^3 sin x ~d x=g(x)+C, where C is the constant of integration. If 8 (g ( π 2 )+g^prime ( π 2 ) )=α π^3+β π^2+γ, α, β, γ ∈ Z, then α+β-γ equals :
JEE Main 2025 — Mathematics Calculus
Let ∫x3sinx dx=g(x)+C, where C is the constant of integration. If 8(g(2π)+g′(2π))=απ3+βπ2+γ,α,β,γ∈Z, then α+β−γ equals :
Held on 23 Jan 2025 · Verified 6 Jul 2026.
Options
- A
48
- B
55
- C
62
- D
47
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