The given integral is of the form $\int \frac{1}{1+u^2} du$, which evaluates to $\tan^{-1}(u)$. We rewrite the integrand as: $$ \int_{0}^{2a} \frac{1}{1+(2x)^2} dx = \frac{\pi}{6} $$
Let $u = 2x$, then $du = 2 dx$ or $dx = \frac{du}{2}$. The limits change: when $x=0$, $u=0$; when $x=2a$, $u=4a$. $$ \frac{1}{2} \int_{0}^{4a} \frac{1}{1+u^2} du = \frac{\pi}{6} $$
Integrating gives: $$ \frac{1}{2} [\tan^{-1}(u)]_{0}^{4a} = \frac{\pi}{6} $$ $$ \tan^{-1}(4a) - \tan^{-1}(0) = \frac{\pi}{3} $$ $$ \tan^{-1}(4a) = \frac{\pi}{3} $$
Taking the tangent of both sides: $$ 4a = \tan\left(\frac{\pi}{3}\right) $$ $$ 4a = \sqrt{3} $$ $$ a = \frac{\sqrt{3}}{4} $$
Final Answer: $\frac{\sqrt{3}}{4}$
AI generated content. Review strictly for academic accuracy.