Calculus – Definite Integration – JEE Main 21 Jan 2026 Shift 1

Solution:

Let $I = \int_{-\pi/6}^{\pi/6} \frac{\pi+4x^{11}}{1-\sin(|x|+\pi/6)} dx$.
Split the integral into two parts:
$$I = \int_{-\pi/6}^{\pi/6} \frac{\pi}{1-\sin(|x|+\pi/6)} dx + \int_{-\pi/6}^{\pi/6} \frac{4x^{11}}{1-\sin(|x|+\pi/6)} dx$$

Let $f(x) = \frac{4x^{11}}{1-\sin(|x|+\pi/6)}$.
Since $|-x| = |x|$ and $(-x)^{11} = -x^{11}$, $f(-x) = -f(x)$.
Thus, $f(x)$ is an odd function, and the second integral is 0.

Now consider the first integral:
$$I = \pi \int_{-\pi/6}^{\pi/6} \frac{1}{1-\sin(|x|+\pi/6)} dx$$
The integrand is an even function.
$$I = 2\pi \int_{0}^{\pi/6} \frac{1}{1-\sin(x+\pi/6)} dx$$

Let $x + \frac{\pi}{6} = t \Rightarrow dx = dt$.
Limits: $x=0 \to t=\pi/6$ and $x=\pi/6 \to t=\pi/3$.
$$I = 2\pi \int_{\pi/6}^{\pi/3} \frac{1}{1-\sin t} dt$$
Multiply numerator and denominator by $(1+\sin t)$:
$$I = 2\pi \int_{\pi/6}^{\pi/3} \frac{1+\sin t}{1-\sin^2 t} dt = 2\pi \int_{\pi/6}^{\pi/3} \frac{1+\sin t}{\cos^2 t} dt$$
$$I = 2\pi \int_{\pi/6}^{\pi/3} (\sec^2 t + \sec t \tan t) dt$$

$$I = 2\pi [\tan t + \sec t]_{\pi/6}^{\pi/3}$$
$$I = 2\pi [(\tan \frac{\pi}{3} + \sec \frac{\pi}{3}) – (\tan \frac{\pi}{6} + \sec \frac{\pi}{6})]$$
$$I = 2\pi [(\sqrt{3} + 2) – (\frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}})]$$
$$I = 2\pi [\sqrt{3} + 2 – \frac{3}{\sqrt{3}}] = 2\pi [\sqrt{3} + 2 – \sqrt{3}]$$
$$I = 2\pi [2] = 4\pi$$

Ans. (2)