`overset(pi)underset(0)int (1)/(1+3^(cosx))` dx is equal to

Options

A. `pi`
B. 0
C. `(pi)/(2)`
D. None of these

Answer

To solve the integral \( I = \int_0^\pi \frac{1}{1 + 3^{\cos x}} \, dx \), we can use a symmetry property of definite integrals. Let's go through the steps:

### Step 1: Define the Integral
Let
\[ I = \int_0^\pi \frac{1}{1 + 3^{\cos x}} \, dx \]

### Step 2: Use the Symmetry Property
We can use the property of definite integrals:
\[ \int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx \]
In our case, \( a = 0 \) and \( b = \pi \). Thus, we have:
\[ I = \int_0^\pi \frac{1}{1 + 3^{\cos(\pi - x)}} \, dx \]

### Step 3: Simplify the Cosine Term
Using the identity \( \cos(\pi - x) = -\cos x \), we can rewrite the integral:
\[ I = \int_0^\pi \frac{1}{1 + 3^{-\cos x}} \, dx \]

### Step 4: Rewrite the Integral
We can rewrite the expression \( 3^{-\cos x} \) as \( \frac{1}{3^{\cos x}} \):
\[ I = \int_0^\pi \frac{1}{1 + \frac{1}{3^{\cos x}}} \, dx \]
This simplifies to:
\[ I = \int_0^\pi \frac{3^{\cos x}}{3^{\cos x} + 1} \, dx \]

### Step 5: Combine Both Forms of the Integral
Now we have two expressions for \( I \):
1. \( I = \int_0^\pi \frac{1}{1 + 3^{\cos x}} \, dx \)
2. \( I = \int_0^\pi \frac{3^{\cos x}}{3^{\cos x} + 1} \, dx \)

### Step 6: Add the Two Integrals
Adding these two expressions:
\[ 2I = \int_0^\pi \left( \frac{1}{1 + 3^{\cos x}} + \frac{3^{\cos x}}{3^{\cos x} + 1} \right) \, dx \]
This simplifies to:
\[ 2I = \int_0^\pi 1 \, dx \]

### Step 7: Evaluate the Integral
The integral of 1 from 0 to \( \pi \) is simply:
\[ 2I = \pi \]

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

### Final Answer
Thus, the value of the integral is:
\[ \int_0^\pi \frac{1}{1 + 3^{\cos x}} \, dx = \frac{\pi}{2} \]

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