Integrate with sin cos 4

$= \displaystyle \int^{\frac{\pi}{6}}_{0} \sqrt{\frac{1+2 \cos 2x}{\csc x}} \sin (6x) \text{ d}x$ $= \displaystyle \int^{\frac{\pi}{6}}_{0} \sqrt{\sin x + 2 \sin x\left(1-2\sin^{2}x\right) } \sin (6x) \text{ d}x$ $= \displaystyle \int^{\frac{\pi}{6}}_{0} \sqrt{3\sin x - 4 \sin^{3}x} \sin (6x) \text{ d}x$ $= \displaystyle \int^{\frac{\pi}{6}}_{0} \sqrt{\sin(3x)} \sin (6x) \text{ d}x$ $= \displaystyle \int^{\frac{\pi}{6}}_{0} \sqrt{\sin(3x)} 2\sin (3x)\cos(3x) \text{ d}x$

Let $u=\sin(3x); \dfrac{\text{d}u}{3 \cos (3x)}= \text{ d}x$

$= \displaystyle \dfrac{2}{3} \int^{1}_{0} \sqrt{u} u \text{ d}u$ $= \displaystyle \dfrac{2}{3} \int^{1}_{0} u^{\frac{3}{2}} \text{ d}u$ $= \displaystyle \left.\begin{matrix} \dfrac{2}{3} \cdot \dfrac{2}{5} u^{\frac{5}{2}} \end{matrix}\right|_{0}^{1}$ $= \boxed{\dfrac{4}{15} }$

or

$\boxed{ \approx .266 }$