`int_(-pi//2)^(pi//2) (cos x dx)/(1+ e^(x))=`

Options

A. 1
B. 2
C. log 2
D. none

Answer

To solve the integral

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

we will use a property of definite integrals.

### Step 1: Define the integral
Let

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

### Step 2: Use the property of definite integrals
We can use the property of definite integrals that states:

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

In our case, \( a = -\frac{\pi}{2} \) and \( b = \frac{\pi}{2} \), so \( a + b = 0 \). Thus, we have:

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

### Step 3: Simplify the integral
Since \( \cos(-x) = \cos x \), we can rewrite the integral as:

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

Now, we can rewrite the denominator:

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

Thus, we have:

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

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

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

Adding these two equations gives:

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

### Step 5: Combine the integrands
The integrands can be combined:

\[ \frac{\cos x}{1 + e^x} + \frac{\cos x \cdot e^x}{e^x + 1} = \frac{\cos x (1 + e^x)}{1 + e^x} = \cos x. \]

Thus, we have:

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

### Step 6: Evaluate the integral
Now, we can evaluate the integral:

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

Thus,

\[ 2I = 2 \implies I = 1. \]

### Final Answer
The value of the integral is

\[ \boxed{1}. \]