Evaluate the following : `int_(0)^(pi)(dx)/(1+sinx)`

Explanation

To evaluate the integral

\[ I = \int_{0}^{\pi} \frac{dx}{1 + \sin x}, \]

we will use a technique that involves multiplying the integrand by a suitable form of 1.

### Step 1: Multiply by a conjugate
We multiply the integrand by

\[ \frac{1 - \sin x}{1 - \sin x} \]

to simplify the expression. This gives us:

\[ I = \int_{0}^{\pi} \frac{1 - \sin x}{(1 + \sin x)(1 - \sin x)} \, dx. \]

### Step 2: Simplify the denominator
The denominator simplifies as follows:

\[ (1 + \sin x)(1 - \sin x) = 1 - \sin^2 x = \cos^2 x. \]

Thus, we can rewrite the integral as:

\[ I = \int_{0}^{\pi} \frac{1 - \sin x}{\cos^2 x} \, dx. \]

### Step 3: Split the integral
We can separate the integral into two parts:

\[ I = \int_{0}^{\pi} \frac{1}{\cos^2 x} \, dx - \int_{0}^{\pi} \frac{\sin x}{\cos^2 x} \, dx. \]

### Step 4: Recognize the integrals
The first integral is:

\[ \int_{0}^{\pi} \sec^2 x \, dx, \]

and the second integral can be rewritten as:

\[ \int_{0}^{\pi} \sec x \tan x \, dx. \]

### Step 5: Evaluate the first integral
The integral of \(\sec^2 x\) is:

\[ \tan x \bigg|_{0}^{\pi} = \tan(\pi) - \tan(0) = 0 - 0 = 0. \]

### Step 6: Evaluate the second integral
The integral of \(\sec x \tan x\) is:

\[ \sec x \bigg|_{0}^{\pi} = \sec(\pi) - \sec(0) = -1 - 1 = -2. \]

### Step 7: Combine the results
Now, we can combine the results of both integrals:

\[ I = 0 - (-2) = 2. \]

Thus, the value of the integral is

\[ \boxed{2}. \]