The value of the integral ∫_ (π)/(24) ^ (5 π)/(24) ~d x 1+ [3] 2 x is :

Correct answer: B. (π)/(12) Let I = ∫_ π/24 ^ 5π/24 dx 1+ [3] 2x . Apply King's rule ( x → π/4 - x ): I = ∫_ π/24 ^ 5π/24 dx 1+ [3] 2x = ∫_ π/24 ^ 5π/24 [3] 2x 1+ [3] 2x dx . Adding: 2I = ∫_ π/24 ^ 5π/24 dx = (5π)/(24) - (π)/(24) = (π)/(6) . I = (π)/(12) .

JEE Main 202623 JanuaryMorning Shift

Exam: JEE Main
Year: 2026
Date: 23 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Definite Integration

JEE Main 2026 · 23 January · Morning Shift · Mathematics · Definite Integration

The value of the integral ∫_ (π)/(24) ^ (5 π)/(24) ~d x 1+ [3] 2 x...

The value of the integral

\int_{\frac{\pi}{24}}^{\frac{5 \pi}{24}} \frac{\mathrm{~d} x}{1+\sqrt[3]{\tan 2 x}}

is :

Options

A $\frac{\pi}{3}$
B $\frac{\pi}{12}$
C $\frac{\pi}{18}$
D $\frac{\pi}{6}$

Correct answer

B. $\frac{\pi}{12}$

Step-by-step 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}

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