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

Options

A. 1
B. 0
C. -1
D. none of these

Answer

To solve the integral

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

we can use the property of definite integrals which states that

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

### Step 1: Identify \( a \) and \( b \)

In our case, \( a = -\frac{\pi}{2} \) and \( b = \frac{\pi}{2} \). Therefore, we calculate \( a + b \):

\[ a + b = -\frac{\pi}{2} + \frac{\pi}{2} = 0. \]

### Step 2: Substitute into the integral

Now, we substitute \( a + b - x \) into the integral:

\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos(0 - x)}{1 + e^{-x}} \, dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos(-x)}{1 + e^{-x}} \, dx. \]

Since \( \cos(-x) = \cos(x) \), we have:

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

### Step 3: Rewrite the integral

Now, we can rewrite the term \( \frac{1}{1 + e^{-x}} \):

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

Thus, we can express \( I \) as:

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

### Step 4: Combine the two integrals

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

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

Adding these two integrals gives:

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

### Step 5: Simplify the integrand

The integrand simplifies to:

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

Thus, we have:

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

### Step 6: Evaluate the integral

The integral of \( \cos x \) is:

\[ \int \cos x \, dx = \sin x. \]

Evaluating from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \):

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

### Step 7: Solve for \( I \)

Now we can solve for \( I \):

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

### Final Answer

Thus, the value of the integral is:

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