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

JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Indefinite Integrals

Let

\int x^{3} sin xdx = g (x) + C

, where C is the constant of integration.
If

8 (g (\frac{π}{2}) + g^{'} (\frac{π}{2})) = α π^{3} + β π^{2} + γ, α, β, γ \in Z

,
Then

α + β - γ

equals :

Answer: (A)

Integrate by parts repeatedly:

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

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

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

x = \frac{π}{2}

(where

cos \frac{π}{2} = 0, sin \frac{π}{2} = 1

g (\frac{π}{2}) = \frac{3 π ^{2}}{4} - 6

Since

g

is an antiderivative of

x^{3} sin x

g^{'} (x) = x^{3} sin x

, so

g^{'} (\frac{π}{2}) = (\frac{π}{2})^{3} \cdot 1 = \frac{π ^{3}}{8}

Then

8 (g (\frac{π}{2}) + g^{'} (\frac{π}{2})) = 8 (\frac{3 π ^{2}}{4} - 6 + \frac{π ^{3}}{8}) = π^{3} + 6 π^{2} - 48

Comparing with

α π^{3} + β π^{2} + γ

α = 1, β = 6, γ = - 48

α + β - γ = 1 + 6 + 48 = 55

JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Single correct option