The value of `int_-pi^pi cos^2x/[1+a^x].dx`,a>0 is

Explanation

To solve the integral

\[ I = \int_{-\pi}^{\pi} \frac{\cos^2 x}{1 + a^x} \, dx \]

where \( a > 0 \), we will follow these steps:

### Step 1: Set up the integral
We start with the integral defined as:

\[ I = \int_{-\pi}^{\pi} \frac{\cos^2 x}{1 + a^x} \, dx \]

### Step 2: Use the property of definite integrals
Next, we will use the substitution \( x = -x \). Under this substitution, the limits of integration remain the same, and we have:

\[ I = \int_{-\pi}^{\pi} \frac{\cos^2(-x)}{1 + a^{-x}} \, (-dx) \]

Since \(\cos(-x) = \cos x\), we can simplify this to:

\[ I = \int_{-\pi}^{\pi} \frac{\cos^2 x}{1 + \frac{1}{a^x}} \, dx \]

### Step 3: Simplify the integral
Now, simplifying the denominator:

\[ 1 + a^{-x} = \frac{a^x + 1}{a^x} \]

Thus, we can rewrite the integral as:

\[ I = \int_{-\pi}^{\pi} \frac{\cos^2 x \cdot a^x}{a^x + 1} \, dx \]

### Step 4: Combine the two expressions for I
Now we have two expressions for \( I \):

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

Adding these two equations, we get:

\[ 2I = \int_{-\pi}^{\pi} \left( \frac{\cos^2 x}{1 + a^x} + \frac{a^x \cos^2 x}{a^x + 1} \right) \, dx \]

### Step 5: Simplify the combined expression
Now, we can combine the fractions:

\[ \frac{\cos^2 x}{1 + a^x} + \frac{a^x \cos^2 x}{a^x + 1} = \frac{\cos^2 x (a^x + 1)}{(1 + a^x)(a^x + 1)} = \frac{\cos^2 x (1 + a^x)}{(1 + a^x)} = \cos^2 x \]

Thus, we have:

\[ 2I = \int_{-\pi}^{\pi} \cos^2 x \, dx \]

### Step 6: Evaluate the integral of cos²
Now we can evaluate the integral:

\[ \int_{-\pi}^{\pi} \cos^2 x \, dx = 2 \int_{0}^{\pi} \cos^2 x \, dx \]

Using the identity \(\cos^2 x = \frac{1 + \cos 2x}{2}\):

\[ \int_{0}^{\pi} \cos^2 x \, dx = \frac{1}{2} \int_{0}^{\pi} (1 + \cos 2x) \, dx = \frac{1}{2} \left[ x + \frac{\sin 2x}{2} \right]_{0}^{\pi} \]

Evaluating this gives:

\[ = \frac{1}{2} \left[ \pi + 0 - 0 \right] = \frac{\pi}{2} \]

Thus,

\[ \int_{-\pi}^{\pi} \cos^2 x \, dx = 2 \cdot \frac{\pi}{2} = \pi \]

### Step 7: Solve for I
Substituting back, we have:

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

### Final Answer
Thus, the value of the integral is:

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

---