Answer
$\frac{3-\sqrt 3}{2}$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /6} {\sqrt {1 + \sin 2\theta } } d\theta \cr
& {\text{Rationalizing the integrand}} \cr
& = \int {\sqrt {1 + \sin 2\theta } \times \frac{{\sqrt {1 - \sin 2\theta } }}{{\sqrt {1 - \sin 2\theta } }}} d\theta \cr
& = \int {\frac{{\sqrt {1 - {{\sin }^2}2\theta } }}{{\sqrt {1 - \sin 2\theta } }}} d\theta \cr
& {\text{Using the pythagorean identity co}}{{\text{s}}^2}x = 1 - {\sin ^2}x \cr
& = \int {\frac{{\sqrt {{{\cos }^2}2\theta } }}{{\sqrt {1 - \sin 2\theta } }}} d\theta \cr
& = \int {\frac{{\cos 2\theta }}{{\sqrt {1 - \sin 2\theta } }}} d\theta \cr
& {\text{Rewrite the integrand}} \cr
& = -\int {\frac{{ - 2\cos 2\theta }}{{2\sqrt {1 - \sin 2\theta } }}} d\theta \cr
& {\text{Integrating}} \cr
& {\text{ = }} - \left[ {{{\left( {1 - \sin 2\theta } \right)}^{1/2}}} \right] + C \cr
& = -\sqrt {1 - \sin 2\theta } + C \cr
& {\text{Therefore}} \cr
& \int_0^{\pi /6} {\sqrt {1 + \sin 2\theta } } d\theta = - \left[ {\sqrt {1 - \sin 2\theta } } \right]_0^{\pi /6} \cr
& = - \left[ {\sqrt {1 - \sin 2\left( {\frac{\pi }{6}} \right)} - \sqrt {1 - \sin 2\left( 0 \right)} } \right] \cr
& = 1-\sqrt {1 - \frac{{\sqrt 3 }}{2}} \cr
& = 1 - \sqrt { \frac{{2-\sqrt 3 }}{2}} \cr
& = 1 - \sqrt { \frac{{4-2\sqrt 3 }}{4}} \cr
& = 1 - \sqrt { \frac{{(\sqrt 3-1)^2 }}{4}} \cr
&=1-\frac{\sqrt 3-1}{2}\cr
&=\frac{3-\sqrt 3}{2}} $$
