`int_(0)^(pi)(1)/(1+sin x)dx` is equal to

Options

A. 1
B. 2
C. -1
D. -2

Explanation

To solve the integral \( I = \int_{0}^{\pi} \frac{1}{1 + \sin x} \, dx \), we can use a substitution and some properties of trigonometric functions. Here’s a step-by-step solution:

### Step 1: Use the identity for \( \sin x \)
We can express \( \sin x \) in terms of \( \tan \) using the half-angle identity:
\[ \sin x = \frac{2 \tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \]
Thus,
\[ 1 + \sin x = 1 + \frac{2 \tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} = \frac{(1 + \tan^2 \frac{x}{2}) + 2 \tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} = \frac{1 + 2 \tan \frac{x}{2} + \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \]
This simplifies to:
\[ 1 + \sin x = \frac{(1 + \tan \frac{x}{2})^2}{1 + \tan^2 \frac{x}{2}} \]

### Step 2: Substitute \( t = \tan \frac{x}{2} \)
Let \( t = \tan \frac{x}{2} \). Then, the differential \( dx \) can be expressed as:
\[ dx = \frac{2}{1 + t^2} \, dt \]
Also, the limits change as follows:
- When \( x = 0 \), \( t = \tan(0) = 0 \)
- When \( x = \pi \), \( t = \tan\left(\frac{\pi}{2}\right) = \infty \)

### Step 3: Rewrite the integral
Now, substituting these into the integral:
\[ I = \int_{0}^{\infty} \frac{1}{\frac{(1 + t)^2}{1 + t^2}} \cdot \frac{2}{1 + t^2} \, dt \]
This simplifies to:
\[ I = \int_{0}^{\infty} \frac{2(1 + t^2)}{(1 + t)^2} \cdot \frac{1}{1 + t^2} \, dt = \int_{0}^{\infty} \frac{2}{(1 + t)^2} \, dt \]

### Step 4: Evaluate the integral
Now, we can evaluate the integral:
\[ I = 2 \int_{0}^{\infty} \frac{1}{(1 + t)^2} \, dt \]
The integral \( \int \frac{1}{(1 + t)^2} \, dt \) can be computed as:
\[ \int \frac{1}{(1 + t)^2} \, dt = -\frac{1}{1 + t} + C \]
Evaluating from \( 0 \) to \( \infty \):
\[ \left[-\frac{1}{1 + t}\right]_{0}^{\infty} = \left(0 - (-1)\right) = 1 \]
Thus,
\[ I = 2 \cdot 1 = 2 \]

### Final Answer
The value of the integral is:
\[ \int_{0}^{\pi} \frac{1}{1 + \sin x} \, dx = 2 \]