\(\frac{\pi}{12}\)
\(\frac{\pi}{6}\)
To evaluate the integral \(\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1 + \tan^{18}x}\), we employ the symmetry property of definite integrals. The integrand is \(f(x) = \frac{1}{1 + \tan^{18}x}\). We seek simplifying properties.
Applying the substitution \(x = \frac{\pi}{3} - t\), we get \(dx = -dt\). The limits of integration change as follows: when \(x = \frac{\pi}{6}, t = \frac{\pi}{6}\) and when \(x = \frac{\pi}{3}, t = 0\). The integral becomes:
\[\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1 + \tan^{18}x} = -\int_{\frac{\pi}{6}}^{0} \frac{dt}{1 + \tan^{18}(\frac{\pi}{3} - t)}\]
Reversing the integration limits yields:
\[\int_{0}^{\frac{\pi}{6}} \frac{dt}{1 + \tan^{18}(\frac{\pi}{3} - t)} = \int_{0}^{\frac{\pi}{6}} \frac{dt}{1 + \left(\frac{1}{\tan t}\right)^{18}}\]
This simplifies to:
\[\int_{0}^{\frac{\pi}{6}} \frac{\tan^{18}t}{1 + \tan^{18}t} dt\]
Now, consider the original integral in relation to this result:
\[\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1 + \tan^{18}x} = \int_{0}^{\frac{\pi}{6}} \frac{dt}{1 + \tan^{18}t} + \int_{0}^{\frac{\pi}{6}} \frac{\tan^{18}t}{1 + \tan^{18}t} dt\]
Adding these two integrals gives:
\[\int_{0}^{\frac{\pi}{6}} \left( \frac{1}{1 + \tan^{18}t} + \frac{\tan^{18}t}{1 + \tan^{18}t} \right) dt = \int_{0}^{\frac{\pi}{6}} dt = \frac{\pi}{6}\]
Therefore, the original integral is half of this sum:
\[\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1 + \tan^{18}x} = \frac{1}{2} \cdot \frac{\pi}{6} = \frac{\pi}{12}\]
The value of the integral is \(\frac{\pi}{12}\).