Evaluate: `int_0^(pi//2) e^x (sinx-cosx) dx`

Answer

To evaluate the integral \( I = \int_0^{\frac{\pi}{2}} e^x (\sin x - \cos x) \, dx \), we can break it down into two separate integrals:

\[ I = \int_0^{\frac{\pi}{2}} e^x \sin x \, dx - \int_0^{\frac{\pi}{2}} e^x \cos x \, dx \]

Let’s denote:
\[ I_1 = \int_0^{\frac{\pi}{2}} e^x \sin x \, dx \]
\[ I_2 = \int_0^{\frac{\pi}{2}} e^x \cos x \, dx \]

Thus, we have:
\[ I = I_1 - I_2 \]

### Step 1: Calculate \( I_1 \)

To calculate \( I_1 \), 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:
\[ I_1 = \left[ e^x \sin x \right]_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} e^x \cos x \, dx \]

Calculating the boundary term:
\[ \left[ e^x \sin x \right]_0^{\frac{\pi}{2}} = e^{\frac{\pi}{2}} \sin\left(\frac{\pi}{2}\right) - e^0 \sin(0) = e^{\frac{\pi}{2}} \cdot 1 - 0 = e^{\frac{\pi}{2}} \]

Thus,
\[ I_1 = e^{\frac{\pi}{2}} - I_2 \]

### Step 2: Calculate \( I_2 \)

Now, we can calculate \( I_2 \) using integration by parts similarly:
Let:
- \( u = \cos x \) and \( dv = e^x \, dx \)
- Then, \( du = -\sin x \, dx \) and \( v = e^x \)

Using integration by parts:
\[ I_2 = \left[ e^x \cos x \right]_0^{\frac{\pi}{2}} + \int_0^{\frac{\pi}{2}} e^x \sin x \, dx \]

Calculating the boundary term:
\[ \left[ e^x \cos x \right]_0^{\frac{\pi}{2}} = e^{\frac{\pi}{2}} \cos\left(\frac{\pi}{2}\right) - e^0 \cos(0) = e^{\frac{\pi}{2}} \cdot 0 - 1 = -1 \]

Thus,
\[ I_2 = -1 + I_1 \]

### Step 3: Substitute \( I_2 \) in \( I_1 \)

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:
\[ I_1 = e^{\frac{\pi}{2}} - (-1 + I_1) \]
\[ I_1 = e^{\frac{\pi}{2}} + 1 - I_1 \]
\[ 2I_1 = e^{\frac{\pi}{2}} + 1 \]
\[ I_1 = \frac{e^{\frac{\pi}{2}} + 1}{2} \]

### Step 4: Calculate \( I_2 \)

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: Calculate \( I \)

Finally, substituting \( I_1 \) and \( I_2 \) back into the expression for \( I \):
\[ I = I_1 - I_2 = \left( \frac{e^{\frac{\pi}{2}} + 1}{2} \right) - \left( \frac{e^{\frac{\pi}{2}} - 1}{2} \right) \]
\[ I = \frac{(e^{\frac{\pi}{2}} + 1) - (e^{\frac{\pi}{2}} - 1)}{2} = \frac{2}{2} = 1 \]

Thus, the final result is:
\[ \boxed{1} \]