The value of `int_(-pi//2)^(pi//2) log {(2-sin theta)/(2+sin theta}d theta`

Options

A. 0
B. 1
C. 2
D. None of these

Answer

To solve the integral

\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \log \left( \frac{2 - \sin \theta}{2 + \sin \theta} \right) d\theta, \]

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

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

### Step 1: Define the function
Let

\[ f(\theta) = \log \left( \frac{2 - \sin \theta}{2 + \sin \theta} \right). \]

### Step 2: Find \( f(-\theta) \)
Now, we will evaluate \( f(-\theta) \):

\[ f(-\theta) = \log \left( \frac{2 - \sin(-\theta)}{2 + \sin(-\theta)} \right). \]

Using the property of sine, \( \sin(-\theta) = -\sin(\theta) \):

\[ f(-\theta) = \log \left( \frac{2 + \sin \theta}{2 - \sin \theta} \right). \]

### Step 3: Simplify \( f(-\theta) \)
Using the logarithmic property \( \log \left( \frac{a}{b} \right) = -\log \left( \frac{b}{a} \right) \):

\[ f(-\theta) = -\log \left( \frac{2 - \sin \theta}{2 + \sin \theta} \right) = -f(\theta). \]

### Step 4: Apply the property of definite integrals
Since we have shown that \( f(-\theta) = -f(\theta) \), we can apply the property:

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

### Conclusion
Thus, the value of the integral is

\[ \boxed{0}. \]