The value of the integral `int_(0)^(pi) (1)/(e^(cosx)+1)dx`, is

Options

A. `pi`
B. 0
C. `2pi`
D. `(pi)/(2)`

Explanation

To solve the integral \( I = \int_{0}^{\pi} \frac{1}{e^{\cos x} + 1} \, dx \), we can use the property of definite integrals. Here’s a step-by-step solution:

### Step 1: Define the integral
Let
\[ I = \int_{0}^{\pi} \frac{1}{e^{\cos x} + 1} \, dx \]

### Step 2: Apply the property of definite integrals
Using the property that
\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx, \]
we can rewrite our integral as:
\[ I = \int_{0}^{\pi} \frac{1}{e^{\cos(\pi - x)} + 1} \, dx \]

### Step 3: Simplify the expression
We know that \( \cos(\pi - x) = -\cos x \). Therefore, we can substitute this into our integral:
\[ I = \int_{0}^{\pi} \frac{1}{e^{-\cos x} + 1} \, dx \]

### Step 4: Rewrite the integral
We can rewrite \( e^{-\cos x} \) as \( \frac{1}{e^{\cos x}} \):
\[ I = \int_{0}^{\pi} \frac{1}{\frac{1}{e^{\cos x}} + 1} \, dx = \int_{0}^{\pi} \frac{e^{\cos x}}{1 + e^{\cos x}} \, dx \]

### Step 5: Add the two integrals
Now we have two expressions for \( I \):
1. \( I = \int_{0}^{\pi} \frac{1}{e^{\cos x} + 1} \, dx \)
2. \( I = \int_{0}^{\pi} \frac{e^{\cos x}}{1 + e^{\cos x}} \, dx \)

Adding these two equations gives:
\[ 2I = \int_{0}^{\pi} \left( \frac{1}{e^{\cos x} + 1} + \frac{e^{\cos x}}{1 + e^{\cos x}} \right) \, dx \]

### Step 6: Simplify the sum
The right-hand side can be simplified:
\[ \frac{1}{e^{\cos x} + 1} + \frac{e^{\cos x}}{1 + e^{\cos x}} = \frac{1 + e^{\cos x}}{e^{\cos x} + 1} = 1 \]
Thus,
\[ 2I = \int_{0}^{\pi} 1 \, dx \]

### Step 7: Calculate the integral
The integral of 1 from 0 to \( \pi \) is simply:
\[ \int_{0}^{\pi} 1 \, dx = \pi \]
So,
\[ 2I = \pi \]

### Step 8: Solve for \( I \)
Dividing both sides by 2 gives:
\[ I = \frac{\pi}{2} \]

### Final Answer
Thus, the value of the integral is:
\[ \boxed{\frac{\pi}{2}} \]