`int_(0)^(pi//2) e^x(sinx+cosx) dx `

Answer

To solve the integral \( I = \int_{0}^{\frac{\pi}{2}} e^x (\sin x + \cos x) \, dx \), we will use integration by parts and properties of definite integrals.

### Step 1: Split the integral
We can split the integral into two parts:
\[ I = \int_{0}^{\frac{\pi}{2}} e^x \sin x \, dx + \int_{0}^{\frac{\pi}{2}} e^x \cos x \, dx \]
Let:
\[ I_1 = \int_{0}^{\frac{\pi}{2}} e^x \sin x \, dx \]
\[ I_2 = \int_{0}^{\frac{\pi}{2}} e^x \cos x \, dx \]
Thus, \( I = I_1 + I_2 \).

### Step 2: Solve \( I_1 \) using integration by parts
For \( I_1 \), we will use integration by parts. Let:
- \( u = \sin x \) and \( dv = e^x \, dx \)
Then:
- \( du = \cos x \, dx \) and \( v = e^x \)

Using integration by parts:
\[ I_1 = \left[ e^x \sin x \right]_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} e^x \cos x \, dx \]
Evaluating the boundary term:
\[ \left[ e^x \sin x \right]_{0}^{\frac{\pi}{2}} = e^{\frac{\pi}{2}} \sin\left(\frac{\pi}{2}\right) - e^0 \sin(0) = e^{\frac{\pi}{2}} - 0 = e^{\frac{\pi}{2}} \]
Thus:
\[ I_1 = e^{\frac{\pi}{2}} - I_2 \]

### Step 3: Solve \( I_2 \) using integration by parts
Now we apply integration by parts to \( I_2 \):
Let:
- \( u = \cos x \) and \( dv = e^x \, dx \)
Then:
- \( du = -\sin x \, dx \) and \( v = e^x \)

Using integration by parts:
\[ I_2 = \left[ e^x \cos x \right]_{0}^{\frac{\pi}{2}} + \int_{0}^{\frac{\pi}{2}} e^x \sin x \, dx \]
Evaluating the boundary term:
\[ \left[ e^x \cos x \right]_{0}^{\frac{\pi}{2}} = e^{\frac{\pi}{2}} \cos\left(\frac{\pi}{2}\right) - e^0 \cos(0) = 0 - 1 = -1 \]
Thus:
\[ I_2 = -1 + I_1 \]

### Step 4: Solve the system of equations
Now we have two equations:
1. \( I_1 = e^{\frac{\pi}{2}} - I_2 \)
2. \( I_2 = -1 + I_1 \)

Substituting \( I_2 \) from the second equation into the first:
\[ I_1 = e^{\frac{\pi}{2}} - (-1 + I_1) \]
This simplifies to:
\[ I_1 = e^{\frac{\pi}{2}} + 1 - I_1 \]
Adding \( I_1 \) to both sides:
\[ 2I_1 = e^{\frac{\pi}{2}} + 1 \]
Thus:
\[ I_1 = \frac{e^{\frac{\pi}{2}} + 1}{2} \]

Now substituting back to find \( I_2 \):
\[ I_2 = -1 + I_1 = -1 + \frac{e^{\frac{\pi}{2}} + 1}{2} = \frac{e^{\frac{\pi}{2}} - 1}{2} \]

### Step 5: Find the final value of \( I \)
Now substituting \( I_1 \) and \( I_2 \) back into \( I \):
\[ I = I_1 + I_2 = \frac{e^{\frac{\pi}{2}} + 1}{2} + \frac{e^{\frac{\pi}{2}} - 1}{2} \]
Combining these:
\[ I = \frac{(e^{\frac{\pi}{2}} + 1) + (e^{\frac{\pi}{2}} - 1)}{2} = \frac{2e^{\frac{\pi}{2}}}{2} = e^{\frac{\pi}{2}} \]

### Final Answer
Thus, the value of the integral is:
\[ I = e^{\frac{\pi}{2}} \]