Let int 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 in Z , Then α+β-γ equals :

Mathematics · Indefinite Integration

JEE Main 2025 — 23 January, Evening Shift — Question 4

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

$8\left(g\left(\frac{\pi}{2}\right)+\mathrm{g}^{\prime}\left(\frac{\pi}{2}\right)\right)=\alpha \pi^{3}+\beta \pi^{2}+\gamma, \alpha, \beta, \gamma \in \mathrm{Z}$ , Then $\alpha+\beta-\gamma$ equals :

Option A:
55
Correct
Option B:
47
Option C:
48
Option D:
62

Answer: A

Step-by-step solution

$\int x^{3} \sin x d x=-x^{3} \cos x+\int 3 x^{2} \cos x d x$

$=-x^{3} \cos x+3 x^{2} \sin x-\int 6 x \sin x d x$

$=-x^{3} \cos x+3 x^{2} \sin x+6 x \cos x-6 \sin x+c$

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

$\mathrm{g}\left(\frac{\pi}{2}\right)=\frac{3 \pi^{2}}{4}-6$

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

$g^{\prime}\left(\frac{\pi}{2}\right)=\frac{\pi^{3}}{8}$

$8\left(g\left(\frac{\pi}{2}\right)+g^{\prime}\left(\frac{\pi}{2}\right)\right)=\pi^{3}+6 \pi^{2}-48$

So $\alpha+\beta-\gamma=55$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 23 January, Evening Shift
Subject: Mathematics
Chapter: Indefinite Integration
Topic: Methods of Indefinite Integration