$\int\limits_{\pi /6}^{\pi /3} {\cfrac{{\sin x + \cos x}}{{\sqrt {\sin 2x} }}dx}$

Let $I = \int\limits_{\pi /6}^{\pi /3} {\cfrac{{\sin x + \cos x}}{{\sqrt {\sin 2x} }}dx} = \int\limits_{\pi /6}^{\pi /3} {\cfrac{{\sin x + \cos x}}{{\sqrt {1 - \left( {1 - \sin 2x} \right)} }}dx}$

$\int\limits_{\pi /6}^{\pi /3} {\cfrac{{\sin x + \cos x}}{{\sqrt {1 - {{\left( {\sin x - \cos x} \right)}^2}} }}dx}$

Put $\sin x - \cos x = t$ $\Rightarrow$ $\left( {\cos x + \sin x} \right)dx = dt$

When $x = \cfrac{\pi }{6},t = \sin \cfrac{\pi }{6} - \cos \cfrac{\pi }{6} = \cfrac{1}{2} - \cfrac{{\sqrt 3 }}{2}$ and

When $x = \cfrac{\pi }{3},t = \sin \cfrac{\pi }{3} - \cos \cfrac{\pi }{3} = \cfrac{{\sqrt 3 }}{2} - \cfrac{1}{2}$

$\therefore$ $I = \int\limits_{\cfrac{1}{2} - \cfrac{{\sqrt 3 }}{2}}^{\cfrac{{\sqrt 3 }}{2} - \cfrac{1}{2}} {\cfrac{{dt}}{{\sqrt {1 - {t^2}} }}} = \left[ {{{\sin }^{ - 1}}t} \right]_{\cfrac{1}{2} - \cfrac{{\sqrt 3 }}{2}}^{\cfrac{{\sqrt 3 }}{2} - \cfrac{1}{2}}$

$= {\sin ^{ - 1}}\left( {\cfrac{{\sqrt 3 }}{2} - \cfrac{1}{2}} \right) - {\sin ^{ - 1}}\left( {\cfrac{1}{2} - \cfrac{{\sqrt 3 }}{2}} \right)$

$= {\sin ^{ - 1}}\left( {\cfrac{{\sqrt 3 }}{2} - \cfrac{1}{2}} \right) + {\sin ^{ - 1}}\left( {\cfrac{{\sqrt 3 }}{2} - \cfrac{1}{2}} \right) = 2{\sin ^{ - 1}}\cfrac{1}{2}\left( {\sqrt 3 - 1} \right)$