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

Answer

To evaluate the integral

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

we can split this 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'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: Evaluate \( I_1 \)

To evaluate \( I_1 \), we will use integration by parts. Let:

- \( u = \sin x \) → \( du = \cos x \, dx \)
- \( dv = e^x \, dx \) → \( v = e^x \)

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:

\[ \left[ e^x \sin x \right]_0^{\frac{\pi}{2}} = e^{\frac{\pi}{2}} \sin\left(\frac{\pi}{2}\right) - e^0 \sin(0) = e^{\frac{\pi}{2}} \cdot 1 - 1 \cdot 0 = e^{\frac{\pi}{2}}. \]

Thus, we have:

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

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

Now, we can evaluate \( I_2 \) using integration by parts as well. Let:

- \( u = \cos x \) → \( du = -\sin x \, dx \)
- \( dv = e^x \, dx \) → \( v = e^x \)

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:

\[ \left[ e^x \cos x \right]_0^{\frac{\pi}{2}} = e^{\frac{\pi}{2}} \cos\left(\frac{\pi}{2}\right) - e^0 \cos(0) = e^{\frac{\pi}{2}} \cdot 0 - 1 \cdot 1 = -1. \]

Thus, we have:

\[ I_2 = -1 + I_1. \]

### Step 3: Substitute back

Now we can substitute \( I_2 \) back into the equation for \( I_1 \):

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

This simplifies to:

\[ I_1 = e^{\frac{\pi}{2}} + 1 - I_1. \]

Adding \( I_1 \) to both sides gives:

\[ 2I_1 = e^{\frac{\pi}{2}} + 1. \]

Thus,

\[ I_1 = \frac{e^{\frac{\pi}{2}} + 1}{2}. \]

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

Now substituting \( I_1 \) back into the equation for \( I_2 \):

\[ I_2 = -1 + \frac{e^{\frac{\pi}{2}} + 1}{2} = \frac{e^{\frac{\pi}{2}} + 1 - 2}{2} = \frac{e^{\frac{\pi}{2}} - 1}{2}. \]

### Step 5: Final Calculation of \( I \)

Now we can find \( I \):

\[ I = I_1 - I_2 = \left( \frac{e^{\frac{\pi}{2}} + 1}{2} \right) - \left( \frac{e^{\frac{\pi}{2}} - 1}{2} \right). \]

This simplifies to:

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

Thus, the final answer is:

\[ \int_0^{\frac{\pi}{2}} e^x (\sin x - \cos x) \, dx = 1. \]