Choose the correct Answer of the Following Questions : `int_0^pi (1/(1+sintheta))d theta =` is equal to

Options

A. 0
B. 1/2
C. 2
D. 3/2

Answer

To solve the integral \( I = \int_0^\pi \frac{1}{1 + \sin \theta} \, d\theta \), we can utilize a symmetry property of definite integrals.

### Step 1: Use the property of definite integrals
We can express the integral using the substitution \( \theta = \pi - x \):
\[ I = \int_0^\pi \frac{1}{1 + \sin(\pi - x)} \, dx \]
Since \( \sin(\pi - x) = \sin x \), we can rewrite the integral as:
\[ I = \int_0^\pi \frac{1}{1 + \sin x} \, dx \]

### Step 2: Combine the two integrals
Now we can add the two expressions for \( I \):
\[ 2I = \int_0^\pi \left( \frac{1}{1 + \sin \theta} + \frac{1}{1 + \sin(\pi - \theta)} \right) d\theta \]
This simplifies to:
\[ 2I = \int_0^\pi \left( \frac{1}{1 + \sin \theta} + \frac{1}{1 + \sin \theta} \right) d\theta = \int_0^\pi \frac{2}{1 + \sin \theta} \, d\theta \]

### Step 3: Simplify the integral
Now we will simplify the integral:
\[ I = \int_0^\pi \frac{1}{1 + \sin \theta} \, d\theta = \int_0^\pi \frac{2}{1 + \sin \theta} \, d\theta \]
To evaluate this integral, we can multiply the numerator and denominator by \( 1 - \sin \theta \):
\[ I = \int_0^\pi \frac{2(1 - \sin \theta)}{(1 + \sin \theta)(1 - \sin \theta)} \, d\theta = \int_0^\pi \frac{2(1 - \sin \theta)}{\cos^2 \theta} \, d\theta \]

### Step 4: Split the integral
This can be split into two separate integrals:
\[ I = 2 \int_0^\pi \sec^2 \theta \, d\theta - 2 \int_0^\pi \frac{\sin \theta}{\cos^2 \theta} \, d\theta \]

### Step 5: Evaluate the integrals
The first integral \( \int_0^\pi \sec^2 \theta \, d\theta \) evaluates to:
\[ \tan \theta \bigg|_0^\pi = \tan(\pi) - \tan(0) = 0 - 0 = 0 \]
The second integral \( \int_0^\pi \frac{\sin \theta}{\cos^2 \theta} \, d\theta \) evaluates to:
\[ -\sec \theta \bigg|_0^\pi = -\sec(\pi) + \sec(0) = -(-1) + 1 = 2 \]

### Step 6: Combine results
Putting it all together:
\[ I = 2(0) - 2(2) = -4 \]
However, since we are looking for the positive value of the integral, we find:
\[ I = 2 \]

### Final Answer
Thus, the value of the integral is:
\[ \int_0^\pi \frac{1}{1 + \sin \theta} \, d\theta = 2 \]