`int_(0)^(pi//2) e^(x) (sin x + cos x) dx`

Answer

To solve the integral

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

we can use the integration formula:

\[ \int e^{x} (f(x) + f'(x)) \, dx = e^{x} f(x) + C, \]

where \( f(x) \) is a function and \( f'(x) \) is its derivative.

### Step 1: Identify \( f(x) \) and \( f'(x) \)

In our case, we can let:

- \( f(x) = \sin x \)
- \( f'(x) = \cos x \)

Thus, we can rewrite our integral as:

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

### Step 2: Apply the formula

According to the formula, we have:

\[ \int e^{x} (\sin x + \cos x) \, dx = e^{x} \sin x + C. \]

### Step 3: Evaluate the definite integral

Now, we need to evaluate this from \( 0 \) to \( \frac{\pi}{2} \):

\[ I = \left[ e^{x} \sin x \right]_{0}^{\frac{\pi}{2}}. \]

Calculating the upper limit:

\[ \text{At } x = \frac{\pi}{2}: \quad e^{\frac{\pi}{2}} \sin\left(\frac{\pi}{2}\right) = e^{\frac{\pi}{2}} \cdot 1 = e^{\frac{\pi}{2}}. \]

Calculating the lower limit:

\[ \text{At } x = 0: \quad e^{0} \sin(0) = 1 \cdot 0 = 0. \]

### Step 4: Combine the results

Thus, we have:

\[ I = e^{\frac{\pi}{2}} - 0 = e^{\frac{\pi}{2}}. \]

### Final Answer

\[ I = e^{\frac{\pi}{2}}. \]