`int_(0)^(2pi) (sin 2 theta)/(a-b cos 2 theta )d theta = `

Options

A. 0
B. 2
C. `pi/4`
D. `pi/2`

Answer

To solve the integral

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

we can use a property of definite integrals.

### Step 1: Use the property of definite integrals

We know that

\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx. \]

In our case, we will substitute \( \theta \) with \( 2\pi - \theta \):

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

### Step 2: Simplify the integral

Now, simplify the integrand:

\[ \sin(2(2\pi - \theta)) = \sin(4\pi - 2\theta) = -\sin(2\theta), \]
and
\[ \cos(2(2\pi - \theta)) = \cos(4\pi - 2\theta) = \cos(2\theta). \]

Thus, we can rewrite the integral as:

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

### Step 3: Combine the two integrals

Now we have two expressions for \( I \):

1. \( I = \int_{0}^{2\pi} \frac{\sin 2\theta}{a - b \cos 2\theta} \, d\theta \)
2. \( I = \int_{0}^{2\pi} \frac{-\sin 2\theta}{a - b \cos 2\theta} \, d\theta \)

Adding these two equations gives:

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

### Step 4: Conclusion

Thus, we find that

\[ 2I = 0 \implies I = 0. \]

Therefore, the value of the integral is

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