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JEE Main 202523 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
23 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Indefinite Integration

JEE Main 2025 · 23 January · Evening Shift · Mathematics · Indefinite Integration

Let ∫ x^3 x ~d x=g(x)+C , where C is the constant of integration....

Let ∫x3sin⁡x dx=g(x)+C\int x^3 \sin x \mathrm{~d} x=g(x)+C, where CC is the constant of integration. If 8(g(π2)+g′(π2))=απ3+βπ2+γ,α,β,γ∈Z8\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 α+β−γ\alpha+\beta-\gamma equals :

Options

  1. A48
  2. B55
  3. C62
  4. D47

Correct answer

B. 55

Step-by-step solution

∫x3sin⁡xdx=−x3cos⁡x+∫3x2cos⁡xdx=−x3cos⁡x+3x2sin⁡x−∫6xsin⁡xdx=−x3cos⁡x+3x2sin⁡x+6xcos⁡x−6sin⁡x+c\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)=−x3cos⁡x+3x2sin⁡x+6xcos⁡x−6sin⁡xg(x)=-x^3 \cos x+3 x^2 \sin x+6 x \cos x-6 \sin xg(π2)=3π24−6 g′(x)=−3x2cos⁡x+x3sin⁡x+6cos⁡x−6cos⁡x g′(π2)=π388( g(π2)+g′(π2))=π3+6π2−48\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}$SoSo\alpha+\beta-\gamma=55$
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