`int_(0)^(pi) (dx)/(1+2^(tan x))`=

Options

A. 0
B. `pi//4`
C. `pi//2`
D. `pi`

Answer

To solve the integral

\[ I = \int_{0}^{\pi} \frac{dx}{1 + 2^{\tan x}}, \]

we will use a property of definite integrals.

### Step 1: Define the integral

Let

\[ I = \int_{0}^{\pi} \frac{dx}{1 + 2^{\tan x}} \tag{1} \]

### Step 2: Change of variable

Using the property of definite integrals, we can change the variable \(x\) to \(\pi - x\):

\[ I = \int_{0}^{\pi} \frac{dx}{1 + 2^{\tan(\pi - x)}} \]

### Step 3: Simplify the expression

Since \(\tan(\pi - x) = -\tan x\), we can rewrite the integral as:

\[ I = \int_{0}^{\pi} \frac{dx}{1 + 2^{-\tan x}} \tag{2} \]

### Step 4: Rewrite the integral

We can rewrite the expression in the integral (2):

\[ I = \int_{0}^{\pi} \frac{dx}{1 + \frac{1}{2^{\tan x}}} = \int_{0}^{\pi} \frac{2^{\tan x}}{2^{\tan x} + 1} dx \]

### Step 5: Combine the two expressions

Now we have two expressions for \(I\):

1. From equation (1):

\[ I = \int_{0}^{\pi} \frac{dx}{1 + 2^{\tan x}} \]

2. From equation (2):

\[ I = \int_{0}^{\pi} \frac{2^{\tan x}}{2^{\tan x} + 1} dx \]

### Step 6: Add the two integrals

Adding these two equations gives:

\[ 2I = \int_{0}^{\pi} \left( \frac{1}{1 + 2^{\tan x}} + \frac{2^{\tan x}}{2^{\tan x} + 1} \right) dx \]

### Step 7: Simplify the integrand

The integrand simplifies to:

\[ \frac{1 + 2^{\tan x}}{1 + 2^{\tan x}} = 1 \]

Thus, we have:

\[ 2I = \int_{0}^{\pi} 1 \, dx \]

### Step 8: Evaluate the integral

Now we can evaluate the integral:

\[ 2I = [x]_{0}^{\pi} = \pi - 0 = \pi \]

### Step 9: Solve for \(I\)

Dividing both sides by 2 gives:

\[ I = \frac{\pi}{2} \]

### Final Result

Thus, the value of the integral is:

\[ \int_{0}^{\pi} \frac{dx}{1 + 2^{\tan x}} = \frac{\pi}{2} \]