`int_(0)^(pi)(e^(cos x))/(e^(cos x)+e^(-cos x))dx=`

Options

A. `pi`
B. `(pi)/(3)`
C. `(pi)/(2)`
D. `(pi)/(4)`

Answer

To solve the integral

\[ I = \int_{0}^{\pi} \frac{e^{\cos x}}{e^{\cos x} + e^{-\cos x}} \, dx, \]

we can use a property of definite integrals that states:

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

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

Let’s rewrite the integral using the property mentioned above. Here, \( a = 0 \) and \( b = \pi \):

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

### Step 2: Simplify the expression

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

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

### Step 3: Combine the two integrals

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

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

Adding these two equations gives:

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

### Step 4: Simplify the sum inside the integral

Notice that:

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

Thus, we have:

\[ 2I = \int_{0}^{\pi} 1 \, dx. \]

### Step 5: Evaluate the integral

The integral of 1 from 0 to \( \pi \) is simply:

\[ \int_{0}^{\pi} 1 \, dx = \pi. \]

So, we have:

\[ 2I = \pi \implies I = \frac{\pi}{2}. \]

### Final Result

The value of the integral is:

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

---