The value of the integral

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

Let $\sin \theta=t, \cos \theta d \theta=d t$

$=\int\left(t^{6}+t^{4}+t^{2}\right) \sqrt{2 t^{4}+3 t^{2}+6} d t=\int\left(t^{5}+t^{3}+t\right) \sqrt{2 t^{6}+3 t^{4}+6 t^{2}} d t$

Let $2 t^{6}+3 t^{4}+6 t^{2}=z$

$12\left(t^{5}+t^{3}+t\right) d t=d z$

$=\frac{1}{12} \int \sqrt{z} d z=\frac{1}{18} z^{3 / 2}+c$

$=\frac{1}{18}\left[\left(2 \sin ^{6} \theta+3 \sin ^{4} \theta+6 \sin ^{2} \theta\right)^{3 / 2}+C\right.$

$=\frac{1}{18}\left[\left(1-\cos ^{2} \theta\right)\left(2\left(1-\cos ^{2} \theta\right)^{2}+3-3 \cos ^{2} \theta+6\right)\right]^{3 / 2}+C$

$=\frac{1}{18}\left[\left(1-\cos ^{2} \theta\right)\left(2 \cos ^{4} \theta-7 \cos ^{2} \theta+11\right)\right]^{3 / 2}+C$

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