The value of the integral 24^ 5 24 dx1 + [3] 2x is: — Solution | JEE Mathematics

MCQMediumJEE 2026Definite Integrals

The value of the integral $\int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} \frac{dx}{1 + \sqrt[3]{\tan 2x}}$ is :

Answer

Correct answer:C

Standard Method

Given:

I = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} \frac{dx}{1 + \sqrt[3]{\tan 2x}}

Find: The value of the integral and the correct option.

Use the property

\int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx

Here,

a = \frac{\pi}{24}, \quad b = \frac{5\pi}{24}, \quad a+b = \frac\pi4

So, replacing $x$ by $\frac\pi4 - x$ ,

I = \int_{\pi/24}^{5\pi/24} \frac{dx}{1 + \sqrt[3]{\tan\left(\frac\pi2 - 2x\right)}}

Since

\tan\left(\frac\pi2 - 2x\right) = \cot 2x

we get

I = \int_{\pi/24}^{5\pi/24} \frac{dx}{1 + \sqrt[3]{\cot 2x}}

Now,

\frac{1}{1 + \sqrt[3]{\cot 2x}} = \frac{\sqrt[3]{\tan 2x}}{1 + \sqrt[3]{\tan 2x}}

Therefore,

I = \int_{\pi/24}^{5\pi/24} \frac{\sqrt[3]{\tan 2x}}{1 + \sqrt[3]{\tan 2x}} \, dx

Adding the two forms of $I$ ,

2I = \int_{\pi/24}^{5\pi/24} \left( \frac{1}{1 + \sqrt[3]{\tan 2x}} + \frac{\sqrt[3]{\tan 2x}}{1 + \sqrt[3]{\tan 2x}} \right) dx

2I = \int_{\pi/24}^{5\pi/24} 1 \, dx = \frac{5\pi}{24} - \frac{\pi}{24} = \frac{4\pi}{24} = \frac\pi6

Hence,

I = \frac\pi{12}

Therefore, the value of the integral is $\frac\pi{12}$ and the correct option is C.

Symmetry Property

Given:

I = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} \frac{dx}{1 + \sqrt[3]{\tan 2x}}

Find: The integral value.

Since the limits satisfy

\frac{\pi}{24} + \frac{5\pi}{24} = \frac\pi4

and the integrand is of the complementary form

\frac{1}{1+u} \quad \text{with} \quad u = \sqrt[3]{\tan 2x}

replacing $x$ by $\frac\pi4 - x$ changes $\tan 2x$ to $\cot 2x$ , so the transformed integrand becomes the complementary part. Hence,

I = \frac{b-a}{2}

with

b-a = \frac{5\pi}{24} - \frac{\pi}{24} = \frac\pi6

Therefore,

I = \frac{1}{2} \cdot \frac\pi6 = \frac\pi{12}

Thus, the correct option is C.

Using the symmetry property with the wrong transformed limit expression. The correct substitution is $x \to a+b-x = \frac\pi4 - x$ because the sum of limits is $\frac\pi4$ , not $\frac\pi2$ . Always compute $a+b$ first.
Failing to convert $\tan\left(\frac\pi2 - 2x\right)$ into $\cot 2x$ . Without this identity, the complementary form of the integrand is missed. Use the co-function identity carefully.
Adding the two integrands incorrectly. After transformation, the sum becomes
$\frac{1}{1+u} + \frac{u}{1+u} = 1$
where $u = \sqrt[3]{\tan 2x}$ . Do not try to add numerators or denominators separately in an invalid way.

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