`int_(0)^(pi) (dx)/(1+ 4^(cos x))=`

Options

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

Answer

To solve the integral

\[ I = \int_{0}^{\pi} \frac{dx}{1 + 4^{\cos x}}, \]

we can use a symmetry property of definite integrals.

### Step 1: Define the integral
Let

\[ I = \int_{0}^{\pi} \frac{dx}{1 + 4^{\cos x}}. \]

### Step 2: Use the property of definite integrals
We can apply the property of definite integrals, which states:

\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx. \]

In our case, we set \( a = \pi \), so we have:

\[ I = \int_{0}^{\pi} \frac{dx}{1 + 4^{\cos(\pi - x)}}. \]

### Step 3: Simplify \( \cos(\pi - x) \)
Using the identity \( \cos(\pi - x) = -\cos x \), we can rewrite the integral:

\[ I = \int_{0}^{\pi} \frac{dx}{1 + 4^{-\cos x}}. \]

### Step 4: Rewrite the integral
We can express \( 4^{-\cos x} \) as \( \frac{1}{4^{\cos x}} \):

\[ I = \int_{0}^{\pi} \frac{dx}{1 + \frac{1}{4^{\cos x}}} = \int_{0}^{\pi} \frac{4^{\cos x}}{4^{\cos x} + 1} \, dx. \]

### Step 5: Combine the two expressions for \( I \)
Now we have two expressions for \( I \):

1. \( I = \int_{0}^{\pi} \frac{dx}{1 + 4^{\cos x}} \)
2. \( I = \int_{0}^{\pi} \frac{4^{\cos x}}{4^{\cos x} + 1} \, dx \)

Adding these two equations gives:

\[ 2I = \int_{0}^{\pi} \left( \frac{1}{1 + 4^{\cos x}} + \frac{4^{\cos x}}{4^{\cos x} + 1} \right) dx. \]

### Step 6: Simplify the integrand
The integrand simplifies as follows:

\[ \frac{1}{1 + 4^{\cos x}} + \frac{4^{\cos x}}{4^{\cos x} + 1} = \frac{1 + 4^{\cos x}}{1 + 4^{\cos x}} = 1. \]

### Step 7: Evaluate the integral
Now we have:

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

### Step 8: Solve for \( I \)
Dividing both sides by 2 gives:

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

Thus, the value of the integral is

\[ \boxed{\frac{\pi}{2}}. \]