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 will use properties of definite integrals.

### Step 1: Define the function
Let

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

### Step 2: Apply the property of definite integrals
We will use the property that if

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

then

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

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

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

Using the properties of sine and cosine, we have:

\[ \sin(2(2\pi - \theta)) = \sin(4\pi - 2\theta) = -\sin(2\theta) \]

and

\[ \cos(2\pi - \theta) = \cos(\theta). \]

Thus, we can write:

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

### Step 4: Conclude the integral value
Since we have shown that

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

it follows that

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

Thus, the value of the integral is

\[ \boxed{0}. \]