The value of the integral
`int (sin theta.sin2theta(sin^(6)theta+sin^(4)theta+sin^(2)theta) sqrt(2sin^(4) theta+3sin^(2)theta+6))/(1-cos2theta)d theta` is :
(where c is a constant of integration)

Options

A. `(1)/(18)[11-18sin^(2)theta+9sin^(4)theta-2sin^(6)theta]^((3)/(2))+c`
B. `(1)/(18)[9-2cos^(6)theta-3cos^(4)theta-6cos^(2)theta]^((3)/(2))+c`
C. `(1)/(18)[9-2sin^(6)theta-3sin^(4)theta-6sin^(2)theta]^((3)/(2))+c`
D. `(1)/(18)[11-18cos^(2)theta+9cos^(4)theta-2cos^(6)theta]^((3)/(2))+c`

Answer

To solve the integral

\[ I = \int \frac{\sin \theta \cdot \sin 2\theta \cdot (\sin^6 \theta + \sin^4 \theta + \sin^2 \theta) \sqrt{2\sin^4 \theta + 3\sin^2 \theta + 6}}{1 - \cos 2\theta} \, d\theta, \]

we will follow these steps:

### Step 1: Simplify the Integral
First, we know that

\[ \sin 2\theta = 2 \sin \theta \cos \theta \quad \text{and} \quad 1 - \cos 2\theta = 2\sin^2 \theta. \]

Substituting these into the integral gives:

\[ I = \int \frac{\sin \theta \cdot (2 \sin \theta \cos \theta) \cdot (\sin^6 \theta + \sin^4 \theta + \sin^2 \theta) \sqrt{2\sin^4 \theta + 3\sin^2 \theta + 6}}{2\sin^2 \theta} \, d\theta. \]

### Step 2: Cancel Terms
The \(2\) in the numerator and denominator cancels out, and we can also cancel one \(\sin \theta\):

\[ I = \int \cos \theta \cdot (\sin^6 \theta + \sin^4 \theta + \sin^2 \theta) \sqrt{2\sin^4 \theta + 3\sin^2 \theta + 6} \, d\theta. \]

### Step 3: Substitute \(t = \sin \theta\)
Let \(t = \sin \theta\), then \(d\theta = \frac{dt}{\cos \theta}\). The integral becomes:

\[ I = \int t^6 + t^4 + t^2 \cdot \sqrt{2t^4 + 3t^2 + 6} \, dt. \]

### Step 4: Simplify the Integral
Now we need to simplify the integral further. We can express the integral as:

\[ I = \int (t^6 + t^4 + t^2) \sqrt{2t^4 + 3t^2 + 6} \, dt. \]

### Step 5: Use a New Variable for the Square Root
Let \(u = 2t^4 + 3t^2 + 6\). Then, we differentiate:

\[ du = (8t^3 + 6t) dt. \]

This means we can express \(dt\) in terms of \(du\):

\[ dt = \frac{du}{8t^3 + 6t}. \]

### Step 6: Rewrite the Integral in Terms of \(u\)
Now we can rewrite the integral in terms of \(u\). The expression for \(t^2\) can be derived from \(u\), and we can substitute back into the integral.

### Step 7: Solve the Integral
After substituting and simplifying, we will find:

\[ I = \frac{1}{18} (2t^6 + 3t^4 + 6t^2)^{3/2} + C. \]

### Step 8: Substitute Back \(t = \sin \theta\)
Finally, we substitute \(t\) back to get:

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

### Final Answer
Thus, the value of the integral is:

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