JEE Main 2026 Jan 23 Shift 1, Mathematics Q10: Some Properties of Definite Integrals

JEE Main2026Jan 23, Shift 1Mathematics

The value of the integral $\int_{π /24}^{5 π /24} \frac{d x}{1 + 3 tan 2 x}$ is :

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

Solution

King's rule. The sum of the limits is $\frac{6 π}{24} = \frac{π}{4}$ . Substituting $x \to \frac{π}{4} - x$ , since $tan (\frac{π}{2} - 2 x) = cot 2 x$ :

$I = \int_{π /24}^{5 π /24} \frac{d x}{1 + 3 cot 2 x} = \int_{π /24}^{5 π /24} \frac{3 tan 2 x}{1 + 3 tan 2 x} d x$

Adding to the original integral, the integrand becomes $1$ :

$2 I = \int_{π /24}^{5 π /24} d x = \frac{π}{6} \Rightarrow I = \frac{π}{12}$ .

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JEE Main 2026 Jan 23 Shift 1, Mathematics Q10: Some Properties of Definite Integrals

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