`overset(pi//2)underset(-pi//2)int (cos x)/(1+e^(x))dx=`

Options

A. 1
B. 2
C. `-1`
D. `-2`

Answer

To solve the integral

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

we can use the property of definite integrals that states:

\[ \int_{-a}^{a} f(x) \, dx = \int_{0}^{a} \left( f(x) + f(-x) \right) \, dx \]

### Step 1: Apply the property of definite integrals

Let \( f(x) = \frac{\cos x}{1 + e^x} \). Then we can express our integral as:

\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} f(x) \, dx = \int_{0}^{\frac{\pi}{2}} \left( f(x) + f(-x) \right) \, dx \]

### Step 2: Calculate \( f(-x) \)

Now, we need to find \( f(-x) \):

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

### Step 3: Combine \( f(x) \) and \( f(-x) \)

Now we add \( f(x) \) and \( f(-x) \):

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

Since both fractions have the same denominator \( 1 + e^x \), we can combine them:

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

### Step 4: Set up the new integral

Now we can rewrite the integral:

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

### Step 5: Evaluate the integral

Now we compute the integral:

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

The integral of \( \cos x \) is \( \sin x \), so we evaluate:

\[ \sin x \bigg|_{0}^{\frac{\pi}{2}} = \sin\left(\frac{\pi}{2}\right) - \sin(0) = 1 - 0 = 1 \]

### Final Answer

Thus, the value of the integral is:

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