An Integral

\ $\begin{aligned} I & = \int_0^\frac \pi 6 f(x) \ dx \\ & = \int_0^\frac \pi 6 \sec x \ dx & \small \color{#3D99F6} \text{Multiply up and down by }\sec x + \tan x \\ & = \int_0^\frac \pi 6 \frac {\sec^2 x + \sec x \tan x}{\sec x + \tan x} dx \\ & = \ln (\sec x + \tan x) \bigg|_0^\frac \pi 6 \\ & = \ln \left(\frac 2{\sqrt 3} + \frac 1{\sqrt 3}\right) - \ln (1+0) \\ & = \ln \sqrt 3 \end{aligned}$

$\implies e^{2a} = e^{2\ln \sqrt 3} = e^{\ln 3} = \boxed{3}$

$I=\displaystyle\int_{0}^{\frac{\pi}{6}} \sec x \, dx$

Let $t=\tan(\frac{x}{2}) \implies \cos x = \frac{1-t^2}{1+t^2} \implies dx=\frac{2 \, dt}{1+t^2}$

Then $I=2\displaystyle\int_{0}^{2-\sqrt{3}} \frac{1+t^2}{(1-t^2)(1+t^2)} \, dt = 2\displaystyle\int_{0}^{2-\sqrt{3}} \frac{dt}{1-t^2}$

Use partial fractions for $I=\displaystyle\int_{0}^{2-\sqrt{3}} \frac{1}{t+1} -\frac{1}{t-1}\,dt$ $\implies I= \displaystyle\left. \ln\displaystyle\left|\frac{t+1}{t-1}\displaystyle\right|\displaystyle\right|_{0}^{2-\sqrt{3}} = \ln\sqrt{3}$

So $2I=\ln3 \implies e^{2I}=e^{\ln3}=\boxed{3}$