The value of `int_(-pi)^pi(cos^2x)/(1+a^x)dx,agt0`, is

Options

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

Answer

To solve the integral

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

we will utilize the property of definite integrals and some trigonometric identities.

### Step 1: Use the property of definite integrals
We know that

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

for any function \( f(x) \). We will apply this property by substituting \( x \) with \( -x \):

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

### Step 2: Simplify the integral
Using the fact that \( \cos(-x) = \cos(x) \), we have:

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

### Step 3: Combine the two integrals
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 gives:

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

### Step 4: Simplify the integrand
We can simplify the integrand:

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

Notice that:

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

Thus, we have:

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

### Step 5: Evaluate the integral
Now we need to evaluate \( \int_{-\pi}^{\pi} \cos^2 x \, dx \). We can use the identity:

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

So,

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

The first integral evaluates to:

\[ \frac{1}{2} \cdot (2\pi) = \pi \]

The second integral, \( \int_{-\pi}^{\pi} \cos(2x) \, dx = 0 \) (since cosine is an even function over a symmetric interval).

Thus,

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

### Step 6: Solve for \( I \)
Now substituting back, we get:

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

### Final Answer
The value of the integral is

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