`int_(pi//2)^(pi//2)(cosx)/(1+e^(x))dx` is equal to

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 a property of definite integrals. Specifically, we will use the substitution \( x = -t \).

### Step 1: Substitute \( x = -t \)

When we make this substitution, the limits of integration change as follows:
- When \( x = -\frac{\pi}{2} \), \( t = \frac{\pi}{2} \)
- When \( x = \frac{\pi}{2} \), \( t = -\frac{\pi}{2} \)

The differential \( dx \) becomes \( -dt \). Thus, we can rewrite the integral:

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

### Step 2: Simplify the integral

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

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

### Step 3: Combine 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 4: Simplify the integrand

The integrand simplifies as follows:

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

Thus, we have:

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

### Step 5: Calculate the integral of \( \cos x \)

The integral of \( \cos x \) from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \) is:

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

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

Now substituting back, we have:

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

Thus, the value of the integral is:

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

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