Solution

$= 2\pi \int_0^{\pi/6} \frac{1}{1 - \sin(x + \pi/6)} dx$ let $x + \frac{\pi}{6} = t \Rightarrow dx = dt$

$= 2\pi \int_{\pi/6}^{\pi/3} \frac{dt}{1 - \sin t}$

$= 2\pi \int_{\pi/6}^{\pi/3} \frac{1 + \sin t}{\cos^2 t} dt$

$= 2\pi \left[ \int_{\pi/6}^{\pi/3} \sec^2 t , dt + \int_{\pi/6}^{\pi/3} \sec t \tan t , dt \right]$

$= 2\pi \left[ (\tan t){\pi/6}^{\pi/3} + (\sec t){\pi/6}^{\pi/3} \right]$

$= 2\pi \left[ \left(\sqrt{3} - \frac{1}{\sqrt{3}}\right) + \left(2 - \frac{2}{\sqrt{3}}\right) \right]$

$= 2\pi \left[ \sqrt{3} + 2 - \sqrt{3} \right] = 4\pi$