Question

$= 2\pi\int_{0}^{\pi/6}\mspace{2mu}\frac{1}{1 - sin\left( x + \frac{\pi}{6} \right)}dx$ let $x + \frac{x}{6} = tdx = dt$

$${= 2\pi\int_{\pi/6}^{\pi/3}\mspace{2mu}\frac{dt}{1 - sint} = 2\pi\int_{\pi/6}^{\pi/3}\mspace{2mu}\frac{1 + sint}{\cos^{2}t}dt }{= 2\pi\left\lbrack \int_{\pi/6}^{\pi/3}\mspace{2mu}\sec^{2}tdt + \int_{\pi/6}^{\pi/3}\mspace{2mu} secttantdt \right\rbrack }{= 2\pi\left\lbrack (tant)_{\pi/6}^{\pi/3} + (sect)_{\pi/6}^{\pi/3} \right\rbrack }{= 2\pi\left\lbrack \left( \sqrt{3} - \frac{1}{\sqrt{3}} \right) + \left( 2 - \frac{2}{\sqrt{3}} \right) \right\rbrack }{= 2\pi\lbrack\sqrt{3} + 2 - \sqrt{3}\rbrack = 4\pi}$$