The value of the integral $$\int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} \frac{dx}{1+\sqrt[3]{\tan2x}}$$ is:

We need to evaluate the integral $$I = \int_{\pi/24}^{5\pi/24} \frac{dx}{1 + \sqrt[3]{\tan 2x}}$$.

King’s rule for definite integrals, namely $$\int_a^b f(x)\,dx = \int_a^b f(a + b - x)\,dx$$, is useful because the substitution $$x \to a + b - x$$ often simplifies the integrand.

Here, since $$a + b = \frac{\pi}{24} + \frac{5\pi}{24} = \frac{\pi}{4}$$, applying King’s rule with the substitution $$x \to \frac{\pi}{4} - x$$ transforms the integral into $$I = \int_{\pi/24}^{5\pi/24} \frac{dx}{1 + \tan^{1/3}\bigl(2(\frac{\pi}{4} - x)\bigr)} = \int_{\pi/24}^{5\pi/24} \frac{dx}{1 + \tan^{1/3}\bigl(\tfrac{\pi}{2} - 2x\bigr)}$$.

Using the identity $$\tan\bigl(\tfrac{\pi}{2} - \theta\bigr) = \cot\theta = \frac{1}{\tan\theta}$$ then gives $$I = \int_{\pi/24}^{5\pi/24} \frac{dx}{1 + \bigl(\tfrac{1}{\tan 2x}\bigr)^{1/3}} = \int_{\pi/24}^{5\pi/24} \frac{dx}{1 + \frac{1}{\tan^{1/3}(2x)}}$$.

Multiplying numerator and denominator by $$\tan^{1/3}(2x)$$ leads to $$I = \int_{\pi/24}^{5\pi/24} \frac{\tan^{1/3}(2x)}{\tan^{1/3}(2x) + 1}\,dx$$.

Adding this expression to the original integral yields $$I + I = \int_{\pi/24}^{5\pi/24} \frac{1}{1 + \tan^{1/3}(2x)}\,dx + \int_{\pi/24}^{5\pi/24} \frac{\tan^{1/3}(2x)}{1 + \tan^{1/3}(2x)}\,dx = \int_{\pi/24}^{5\pi/24} 1\,dx$$.

Consequently, $$2I = \frac{5\pi}{24} - \frac{\pi}{24} = \frac{4\pi}{24} = \frac{\pi}{6}$$, which implies $$I = \frac{\pi}{12}$$.

The correct answer is Option (4): $$\frac{\pi}{12}$$.