Q30 / 75 · JEE Main 2026 (28 January Shift 2) · Mathematics
Let $[\cdot]$ denote the greatest integer function. Then $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\frac{12(3+[x])}{3+[\sin x]+[\cos x]}\right) \mathrm{d} x$ is equal to :
- A$13 \pi+1$
- B$12 \pi+5$
- Correct answerC$11 \pi+2$
- D$15 \pi+4$
Solution
$I = \int_{-\pi/2}^{\pi/2} \dfrac{12(3 + [x]) \, dx}{3 + [\sin x] + [\cos x]}$
$I = \int_{-\pi/2}^{-1} \dfrac{12(1) \, dx}{2} + \int_{-1}^{0} \dfrac{12(2) \, dx}{2} + \int_{0}^{1} \dfrac{12(3) \, dx}{3} + \int_{1}^{\pi/2} \dfrac{12(4) \, dx}{3}$
$I = 6\left(\dfrac{\pi}{2} - 1\right) + 12(0 + 1) + 12(1 - 0) + 16\left(\dfrac{\pi}{2} - 1\right)$
$I = 3\pi - 6 + 12 + 12 + 8\pi - 16$
$I = 11\pi + 2$
$I = \int_{-\pi/2}^{-1} \dfrac{12(1) \, dx}{2} + \int_{-1}^{0} \dfrac{12(2) \, dx}{2} + \int_{0}^{1} \dfrac{12(3) \, dx}{3} + \int_{1}^{\pi/2} \dfrac{12(4) \, dx}{3}$
$I = 6\left(\dfrac{\pi}{2} - 1\right) + 12(0 + 1) + 12(1 - 0) + 16\left(\dfrac{\pi}{2} - 1\right)$
$I = 3\pi - 6 + 12 + 12 + 8\pi - 16$
$I = 11\pi + 2$
Hence, the correct answer is option: 'C'.