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Mathematics · Definite Integration

JEE Main 2026 — 28 January, Evening Shift — Question 13

Let [] denote the greatest integer function. Then ∫−π2π2(12(3+[x])3+[sin⁡x]+[cos⁡x])dx\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\frac{12(3+[x])}{3+[\sin x]+[\cos x]}\right) d x is equal to:

  1. Option A:

    15π+415 \pi+4

  2. Option B:

    11π+211 \pi+2

    Correct
  3. Option C:

    13π+113 \pi+1

  4. Option D:

    12π+512 \pi+5

Answer: B

Step-by-step solution

I=∫−π2π212(3+[x])dx3+[sin⁡x]+[cos⁡x]I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{12(3+[x]) d x}{3+[\sin x]+[\cos x]}

I=∫−π2−112dx2+∫−1012dx2+∫0112dx3+∫1π212dx3\mathrm{I}=\int_{-\frac{\pi}{2}}^{-1} \frac{12\mathrm{dx}}{2}+\int_{-1}^{0} \frac{12\mathrm{dx}}{2}+\int_{0}^{1} \frac{12 \mathrm{dx}}{3}+\int_{1}^{\frac{\pi}{2}} \frac{12 \mathrm{dx}}{3}

I=6(π2−1)+12(0+1)+12(1−0)+16(π2−1)\mathrm{I}=6\left(\frac{\pi}{2}-1\right)+12(0+1)+12(1-0)+16\left(\frac{\pi}{2}-1\right)

I=3π−6+12+12+8π−16\mathrm{I}=3 \pi-6+12+12+8 \pi-16

I=11π+2\mathrm{I}=11 \pi+2

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
28 January, Evening Shift
Subject
Mathematics
Chapter
Definite Integration
Topic
Evaluation of Definite Integrals