The value of the integral

Solution:

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

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

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

Let $\sin \theta=\mathrm{t} \Rightarrow \cos \theta \mathrm{d} \theta=\mathrm{dt}$

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

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

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

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

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

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

$\therefore \quad I=\int\left(u^{2}\right)^{1 / 2} \cdot \frac{2 u d u}{12}$

$=\int \frac{\mathrm{u}^{2}}{6} \mathrm{~d} \mathrm{u}=\frac{\mathrm{u}^{3}}{18}+\mathrm{C}$

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

when $\mathrm{t}=\sin \theta$

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