The value of the integral ∫ sin θ · sin 2 θ sin 6 θ + sin 4 θ + sin 2 θ…

$\Rightarrow I=\int \frac{\sin \theta .2\sin \theta \cos \theta ·{\sin }^2\theta \left({\sin }^4\theta +{\sin }^2\theta +1\right){\left(2{\sin }^4\theta +3{\sin }^2\theta +6\right)}^{1/2}}{2{\sin }^2\theta }d\theta$

$=\int {\sin }^2\theta ·\cos \theta \left({\sin }^4\theta +{\sin }^2\theta +1\right){\left(2{\sin }^4\theta +3{\sin }^2\theta +6\right)}^{1/2} d\theta$

Let $\sin \theta =t\Rightarrow \cos \theta d\theta =dt$

$\therefore I=\int t^2\left(t^4+t^2+1\right){\left(2t^4+3t^2+6\right)}^{1/2}dt$

$=\int \left(t^5+t^3+t\right)t{\left(2t^4+3t^2+6\right)}^{1/2}dt$

$=\int \left(t^5+t^3+t\right){\left(t^2\right)}^{1/2}{\left(2t^4+3t^2+6\right)}^{1/2}dt$

$=\int \left(t^5+t^3+t\right){\left(2t^6+3t^4+6t^2\right)}^{1/2}dt$

Let $2t^6+3t^4+6t^2=u^2$

$\Rightarrow 12\left(t^5+t^3+t\right)dt=2udu$

$\therefore I=\int {\left(u^2\right)}^{1/2}·\frac{2udu}{12}$

$=\int \frac{u^2}{6} du=\frac{u^3}{18}+C$

$=\frac{{\left(2t^6+3t^4+6t^2\right)}^{3/2}}{18}+C$

when $t=\sin \theta$

and $t^2=1-{\cos }^2\theta$ will give

$=\frac{1}{18}{\left[11-18{\cos }^2\theta +9{\cos }^4\theta -2{\cos }^6\theta \right]}^{\frac{3}{2}}+c$