The value of the integral $\int \frac{\sin \theta .\sin 2\theta \left({\sin }^6\theta +{\sin }^4\theta +{\sin }^2\theta \right)\sqrt{2{\sin }^4\theta +3{\sin }^2\theta +6}}{1-\cos 2\theta }d\theta$

The given integral is 

$\int \frac{\sin \theta .\sin 2\theta \left({\sin }^6\theta +{\sin }^4\theta +{\sin }^2\theta \right)\sqrt{2{\sin }^4\theta +3{\sin }^2\theta +6}}{1-\cos 2\theta }d\theta =\int \frac{2{\sin }^2\theta \left({\sin }^6\theta +{\sin }^4\theta +{\sin }^2\theta \right)\sqrt{2{\sin }^4\theta +3{\sin }^2\theta +6}}{2{\sin }^2\theta }\cdot \cos \theta d\theta$

substitute $\sin \theta =t,\cos \theta =dt$

The given integral becomes

$\begin{array}{l}=\int \left(t^6+t^4+t^2\right)\sqrt{2t^4+3t^2+6}dt\\ =\int \left(t^5+t^3+t\right)\sqrt{2t^6+3t^4+6t^2}dt\end{array}$

substitute $2t^6+3t^4+6t^2=y\Rightarrow 12\left(t^5+t^3+t\right)dt=dy$

hence, $\frac{1}{12}\int \sqrt{y}dy=\frac{1}{18}{y}^{\frac{3}{2}}+c=\frac{1}{18}{\left(2t^6+3t^4+6t^2\right)}^{\frac{3}{2}}+c$

$\begin{array}{l}=\frac{1}{18}{\left(2{\sin }^6\theta +3{\sin }^4\theta +6{\sin }^2\theta \right)}^{\frac{3}{2}}+C\\ =\frac{1}{18}{\left[-2{\cos }^6\theta +9{\cos }^4\theta -18{\cos }^2\theta +11\right]}^{\frac{3}{2}}+C\end{array}$