Q28 / 75 · JEE Main 2026 (23 January Shift 1) · Mathematics
The value of the integral $\int_{\frac{\pi}{24}}^{\frac{5 \pi}{24}} \frac{\mathrm{~d} x}{1+\sqrt[3]{\tan 2 x}}$ is :
- A$\frac{\pi}{3}$
- Correct answerB$\frac{\pi}{12}$
- C$\frac{\pi}{18}$
- D$\frac{\pi}{6}$
Solution
Let $I = \int_{\pi/24}^{5\pi/24} \frac{dx}{1+\sqrt[3]{\tan 2x}}$.
Apply King's rule ($x \to \pi/4 - x$):
$I = \int_{\pi/24}^{5\pi/24} \frac{dx}{1+\sqrt[3]{\cot 2x}} = \int_{\pi/24}^{5\pi/24} \frac{\sqrt[3]{\tan 2x}}{1+\sqrt[3]{\tan 2x}}dx$.
Adding: $2I = \int_{\pi/24}^{5\pi/24} dx = \frac{5\pi}{24} - \frac{\pi}{24} = \frac{\pi}{6}$.
$I = \frac{\pi}{12}$.
Apply King's rule ($x \to \pi/4 - x$):
$I = \int_{\pi/24}^{5\pi/24} \frac{dx}{1+\sqrt[3]{\cot 2x}} = \int_{\pi/24}^{5\pi/24} \frac{\sqrt[3]{\tan 2x}}{1+\sqrt[3]{\tan 2x}}dx$.
Adding: $2I = \int_{\pi/24}^{5\pi/24} dx = \frac{5\pi}{24} - \frac{\pi}{24} = \frac{\pi}{6}$.
$I = \frac{\pi}{12}$.
Hence, the correct answer is option: 'B'.