Evaluate :
`int_(0)^(pi//2)e^(x)(sin x- cos x)dx`

Answer

To evaluate the integral

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

we can use integration by parts or recognize a pattern in the integrand. Here, we will use the method of recognizing the derivative of a function.

### Step 1: Rewrite the integral
We can rewrite the integral as follows:

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

Let’s denote the two integrals separately:

\[ I_1 = \int_{0}^{\frac{\pi}{2}} e^x \sin x \, dx, \]
\[ I_2 = \int_{0}^{\frac{\pi}{2}} e^x \cos x \, dx. \]

Thus, we have:

\[ I = I_1 - I_2. \]

### Step 2: Evaluate \( I_1 \)
To evaluate \( I_1 \), we 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:

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

Thus, we have:

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

### Step 3: Evaluate \( I_2 \)
Now we can evaluate \( I_2 \) similarly:

Let:

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

Using integration by parts:

\[ I_2 = \left[ e^x \cos x \right]_{0}^{\frac{\pi}{2}} + \int_{0}^{\frac{\pi}{2}} e^x \sin x \, dx. \]

Evaluating the boundary term:

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

Thus, we have:

\[ I_2 = -1 + I_1. \]

### Step 4: Substitute back
Now we have two equations:

1. \( I_1 = e^{\frac{\pi}{2}} - I_2 \)
2. \( I_2 = -1 + I_1 \)

Substituting the second equation into the first:

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

This simplifies to:

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

Combining like terms gives:

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

Thus,

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

### Step 5: Find \( I_2 \)
Now substituting \( I_1 \) back into the equation for \( I_2 \):

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

### Step 6: Final result for \( I \)
Now substituting \( I_1 \) and \( I_2 \) back into the expression for \( I \):

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

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:

\[ \boxed{1}. \]