The value of the integral
`int_(0)^(2pi)(sin2 theta)/(a-b cos theta)d theta` when `a gt b gt 0`, is

Options

A. 1
B. `pi`
C. `pi//2`
D. 0

Answer

To solve the integral

\[ I = \int_{0}^{2\pi} \frac{\sin 2\theta}{a - b \cos \theta} \, d\theta \]

where \( a > b > 0 \), we can use properties of definite integrals.

### Step 1: Identify the function

We define the function:

\[ f(\theta) = \frac{\sin 2\theta}{a - b \cos \theta} \]

### Step 2: Use the property of definite integrals

We will use the property of definite integrals which states that if \( f(2\pi - x) = -f(x) \), then:

\[ \int_0^{2\pi} f(x) \, dx = 0 \]

### Step 3: Calculate \( f(2\pi - \theta) \)

Now, we compute \( f(2\pi - \theta) \):

\[ f(2\pi - \theta) = \frac{\sin(2(2\pi - \theta))}{a - b \cos(2\pi - \theta)} \]

Using the identities \( \sin(2(2\pi - \theta)) = \sin(4\pi - 2\theta) = -\sin(2\theta) \) and \( \cos(2\pi - \theta) = \cos(\theta) \), we have:

\[ f(2\pi - \theta) = \frac{-\sin 2\theta}{a - b \cos \theta} \]

### Step 4: Relate \( f(2\pi - \theta) \) to \( f(\theta) \)

From the above, we can see that:

\[ f(2\pi - \theta) = -f(\theta) \]

### Step 5: Conclude the integral value

Since \( f(2\pi - \theta) = -f(\theta) \), we can conclude that:

\[ \int_0^{2\pi} f(\theta) \, d\theta = 0 \]

Thus, the value of the integral is:

\[ \int_{0}^{2\pi} \frac{\sin 2\theta}{a - b \cos \theta} \, d\theta = 0 \]

### Final Answer

The value of the integral is:

\[ \boxed{0} \]