Integral Calculus – Definite Integrals – JEE Main 22 Jan 2026 Shift 1

Solution:

Let $I = \int_{-\pi/2}^{\pi/2} \frac{1}{[x]+4} dx$.

We split the definite integral based on the critical points of the greatest integer function $[x]$ within the interval $[-\pi/2, \pi/2]$.
Approximating $\pi/2 \approx 1.57$ and $-\pi/2 \approx -1.57$.
The integer points in $(-1.57, 1.57)$ are $-1, 0, 1$.

We break the integral into four intervals:
1. $-\frac{\pi}{2} \le x < -1 \Rightarrow [x] = -2$ 2. $-1 \le x < 0 \Rightarrow [x] = -1$ 3. $0 \le x < 1 \Rightarrow [x] = 0$ 4. $1 \le x < \frac{\pi}{2} \Rightarrow [x] = 1$
Substitute these values into the integral:
$$ I = \int_{-\pi/2}^{-1} \frac{dx}{-2+4} + \int_{-1}^{0} \frac{dx}{-1+4} + \int_{0}^{1} \frac{dx}{0+4} + \int_{1}^{\pi/2} \frac{dx}{1+4} $$
$$ I = \int_{-\pi/2}^{-1} \frac{1}{2} dx + \int_{-1}^{0} \frac{1}{3} dx + \int_{0}^{1} \frac{1}{4} dx + \int_{1}^{\pi/2} \frac{1}{5} dx $$

Evaluating the integrals:
$$ I = \frac{1}{2}[x]_{-\pi/2}^{-1} + \frac{1}{3}[x]_{-1}^{0} + \frac{1}{4}[x]_{0}^{1} + \frac{1}{5}[x]_{1}^{\pi/2} $$
$$ I = \frac{1}{2}\left(-1 – \left(-\frac{\pi}{2}\right)\right) + \frac{1}{3}(0 – (-1)) + \frac{1}{4}(1 – 0) + \frac{1}{5}\left(\frac{\pi}{2} – 1\right) $$

$$ I = \frac{1}{2}\left(\frac{\pi}{2} – 1\right) + \frac{1}{3}(1) + \frac{1}{4}(1) + \frac{1}{5}\left(\frac{\pi}{2} – 1\right) $$
$$ I = \frac{\pi}{4} – \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{\pi}{10} – \frac{1}{5} $$

Grouping the $\pi$ terms and the constant terms:
$$ I = \left(\frac{\pi}{4} + \frac{\pi}{10}\right) + \left(-\frac{1}{2} + \frac{1}{3} + \frac{1}{4} – \frac{1}{5}\right) $$

$$ = \frac{5\pi + 2\pi}{20} + \frac{-30 + 20 + 15 – 12}{60} $$

$$ I = \frac{7\pi}{20} – \frac{7}{60} $$
$$ I = \frac{21\pi}{60} – \frac{7}{60} $$
$$ I = \frac{7}{60}(3\pi – 1) $$

Ans. (3)