The value of `int_(-pi)^(pi)(sqrt2cosx)/(1+e^(x))dx` is equal to

Options

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

Answer

To solve the integral

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

we will use a property of definite integrals. The property states that:

\[ \int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx. \]

### Step 1: Apply the property to the integral

Let’s denote the integral as \( I \):

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

Now, we will apply the property with \( a = -\pi \) and \( b = \pi \):

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

### Step 2: Simplify the integral

Using the fact that \( \cos(-x) = \cos x \), we can rewrite the integral:

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

Now, we simplify the denominator:

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

Thus, we have:

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

### Step 3: Combine the two expressions for \( I \)

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

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

Let’s add these two equations:

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

### Step 4: Combine the fractions

The common denominator is \( 1 + e^x \):

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

### Step 5: Evaluate the integral

Now we need to evaluate:

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

Since \( \sqrt{2} \) is a constant, we can factor it out:

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

The integral of \( \cos x \) over a symmetric interval around zero is zero:

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

Thus,

\[ 2I = \sqrt{2} \cdot 0 = 0. \]

### Step 6: Solve for \( I \)

Dividing both sides by 2 gives:

\[ I = 0. \]

### Final Answer

The value of the integral is:

\[ \boxed{0}. \]