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Mathematics · Definite Integration

JEE Main 2024 — 29 January, Shift 1 — Question 17

If the value of the integral ∫−π2π2(x2cos⁡x1+πx+1+sin⁡2x1+esin⁡sin2023)dx=π4(π+a)−2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\frac{\mathrm{x}^{2} \cos \mathrm{x}}{1+\pi^{\mathrm{x}}}+\frac{1+\sin ^{2} \mathrm{x}}{1+\mathrm{e}^{\sin \mathrm{sin}^{2023}}}\right) \mathrm{dx}=\frac{\pi}{4}(\pi+\mathrm{a})-2, then the value of a is

  1. Option A:

    3

    Correct
  2. Option B:

    −32-\frac{3}{2}

  3. Option C:

    2

  4. Option D:

    32\frac{3}{2}

Answer: A

Step-by-step solution

I=∫−π/2π/2(x2cos⁡x1+πx+1+sin⁡2x1+esin⁡x2023)dxI=\int_{-\pi / 2}^{\pi / 2}\left(\frac{x^{2} \cos x}{1+\pi^{x}}+\frac{1+\sin ^{2} x}{1+e^{\sin x^{2023}}}\right) d x

I=∫−π/2π/2(x2cos⁡x1+π−x+1+sin⁡2x1+esin⁡(−x)2023)dxI=\int_{-\pi / 2}^{\pi / 2}\left(\frac{x^{2} \cos x}{1+\pi^{-x}}+\frac{1+\sin ^{2} x}{1+e^{\sin (-x)^{2023}}}\right) d x

On Adding, we get 2I=∫−π/2π/2(x2cos⁡x+1+sin⁡2x)dx2 I=\int_{-\pi / 2}^{\pi / 2}\left(x^{2} \cos x+1+\sin ^{2} x\right) d x

On solving I=π24+3π4−2I=\frac{\pi^{2}}{4}+\frac{3 \pi}{4}-2

a=3\mathrm{a}=3

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
29 January, Shift 1
Subject
Mathematics
Chapter
Definite Integration
Topic
Methods of solving definite integrals(kings rule,odd even)

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