The value of the integral  $\int \frac{\sin \theta .\sin 2\theta ({\sin }^6\theta +{\sin }^4\theta +{\sin }^2\theta )\sqrt{2{\sin }^4\theta +3{\sin }^2\theta +6}}{1-\cos 2\theta }d\theta$ is : (where C is a constant of integration)

$$\begin{gathered} I=\int \frac{\sin \theta .\sin 2\theta ({\sin }^6\theta +{\sin }^4\theta +{\sin }^2\theta )\sqrt{2{\sin }^4\theta +3{\sin }^2\theta +6}}{1-\cos 2\theta }d\theta \\ \Rightarrow I=\int \frac{\sin \theta .2\sin \theta \cos \theta .{\sin }^2\theta ({\sin }^4\theta +{\sin }^2\theta +1){(2{\sin }^4\theta +3{\sin }^2\theta +6)}^{\frac{1}{2}}}{2{\sin }^2\theta }d\theta \\ Let \sin \theta =t\Rightarrow \cos \theta d\theta =dt \\ \therefore I=\int t^2(t^4+t^2+1){(2t^4+3t^2+6)}^{\frac{1}{2}}dt \\ =\int (t^5+t^3+t)t{(2t^4+3t^2+6)}^{\frac{1}{2}}dt \\ =\int (t^5+t^3+t){\left(t^2\right)}^{\frac{1}{2}}{(2t^4+3t^2+6)}^{\frac{1}{2}}dt \\ =\int (t^5+t^3+t){(2t^6+3t^4+6t^2)}^{\frac{1}{2}}dt \\ Let, 2t^6+3t^4+6t^2=u^2 \\ \Rightarrow 12(t^5+t^3+t)dt=2udu \\ \therefore I=\int {\left(u^2\right)}^{\frac{1}{2}}.\frac{2udu}{12} \\ =\int \frac{u^2}{6}du=\frac{u^3}{18}+C \\ =\frac{{(2t^6+3t^4+6t^2)}^{\frac{3}{2}}}{18}+C \\ When t = \sin \theta \end{gathered}$$

And  $t^2=1-{\cos }^2\theta$ will give option (4)