Answer
$\displaystyle \frac{\sqrt{2}}{2}$
Work Step by Step
Use the identity $\cos 2x=2\cos^{2}x-1$
$I=\displaystyle \int_{0}^{\pi/6}\sqrt{1+\cos 2x}dx=\int_{0}^{\pi/6}\sqrt{1+(2\cos^{2}x-1)}dx$
$=\displaystyle \int_{0}^{\pi/6}\sqrt{2\cos^{2}x}dx$
$=\displaystyle \sqrt{2}\int_{0}^{\pi/6}\sqrt{\cos^{2}x}dx$
$=\displaystyle \sqrt{2}\int_{0}^{\pi/6}|\cos x|dx$
... cosine is positive on the interval $[0,\pi/6]$
... so we can remove the absolute value brackets
$=\displaystyle \sqrt{2}\int_{0}^{\pi/6}\cos xdx$
$=\sqrt{2}[\sin x]_{0}^{\pi/6}$
$=\displaystyle \sqrt{2}(\frac{1}{2}-0)$
$=\displaystyle \frac{\sqrt{2}}{2}$
