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

We simplify the integrand step by step. Note that $$\sin 2\theta = 2\sin\theta\cos\theta$$ and $$1 - \cos 2\theta = 2\sin^2\theta$$. Substituting these:

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

This simplifies to $$\int \cos\theta \cdot (\sin^6\theta + \sin^4\theta + \sin^2\theta) \cdot \sqrt{2\sin^4\theta + 3\sin^2\theta + 6}\, d\theta$$.

Substituting $$t = \sin\theta$$, so $$dt = \cos\theta\, d\theta$$, the integral becomes $$\int (t^6 + t^4 + t^2) \cdot \sqrt{2t^4 + 3t^2 + 6}\, dt$$.

Let $$w = 2t^6 + 3t^4 + 6t^2$$. Then $$\frac{dw}{dt} = 12t^5 + 12t^3 + 12t = 12t(t^4 + t^2 + 1)$$. Also, $$w = t^2(2t^4 + 3t^2 + 6)$$, so $$\sqrt{2t^4 + 3t^2 + 6} = \frac{\sqrt{w}}{t}$$.

Rewriting the integrand: $$t^2(t^4 + t^2 + 1) \cdot \frac{\sqrt{w}}{t}\, dt = t(t^4 + t^2 + 1)\sqrt{w}\, dt = \frac{\sqrt{w}}{12}\, dw$$.

Integrating: $$\frac{1}{12}\int w^{1/2}\, dw = \frac{1}{12} \cdot \frac{2}{3} w^{3/2} + c = \frac{1}{18} w^{3/2} + c$$.

Substituting back $$w = 2\sin^6\theta + 3\sin^4\theta + 6\sin^2\theta$$. To match Option 3, we convert to cosines using $$\sin^2\theta = 1 - \cos^2\theta$$:

$$2(1 - \cos^2\theta)^3 + 3(1 - \cos^2\theta)^2 + 6(1 - \cos^2\theta) = 11 - 18\cos^2\theta + 9\cos^4\theta - 2\cos^6\theta$$.

Therefore, the integral equals $$\frac{1}{18}\left[11 - 18\cos^2\theta + 9\cos^4\theta - 2\cos^6\theta\right]^{3/2} + c$$.