What is `int_(0)^(pi)e^(x) sin x dx ` equal to ?

Options

A. `(e^(pi) + 1)/(2)`
B. `(e^(pi) - 1)/(2)`
C. `e^(pi) + 1`
D. `(e^(pi) + 1)/(4)`

Explanation

To solve the integral \( I = \int_{0}^{\pi} e^{x} \sin x \, dx \), we will use integration by parts.

### Step 1: Set up the integration by parts
We will use the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
Let:
- \( u = \sin x \) (first function)
- \( dv = e^{x} \, dx \) (second function)

Then we need to find \( du \) and \( v \):
- \( du = \cos x \, dx \)
- \( v = e^{x} \)

### Step 2: Apply integration by parts
Now we apply the integration by parts formula:
\[ I = \left[ \sin x \cdot e^{x} \right]_{0}^{\pi} - \int_{0}^{\pi} e^{x} \cos x \, dx \]

### Step 3: Evaluate the boundary term
Evaluate \( \left[ \sin x \cdot e^{x} \right]_{0}^{\pi} \):
\[ \left[ \sin x \cdot e^{x} \right]_{0}^{\pi} = \sin(\pi) e^{\pi} - \sin(0) e^{0} = 0 - 0 = 0 \]
Thus, we have:
\[ I = 0 - \int_{0}^{\pi} e^{x} \cos x \, dx \]
So,
\[ I = - \int_{0}^{\pi} e^{x} \cos x \, dx \]

### Step 4: Apply integration by parts again
Now we need to evaluate \( \int_{0}^{\pi} e^{x} \cos x \, dx \) using integration by parts again. Let:
- \( u = \cos x \)
- \( dv = e^{x} \, dx \)

Then we find:
- \( du = -\sin x \, dx \)
- \( v = e^{x} \)

Applying integration by parts again:
\[ \int e^{x} \cos x \, dx = \left[ \cos x \cdot e^{x} \right]_{0}^{\pi} - \int_{0}^{\pi} e^{x} (-\sin x) \, dx \]
This simplifies to:
\[ \int e^{x} \cos x \, dx = \left[ \cos x \cdot e^{x} \right]_{0}^{\pi} + \int_{0}^{\pi} e^{x} \sin x \, dx \]

### Step 5: Evaluate the boundary term for \( \cos x \)
Evaluate \( \left[ \cos x \cdot e^{x} \right]_{0}^{\pi} \):
\[ \left[ \cos x \cdot e^{x} \right]_{0}^{\pi} = \cos(\pi) e^{\pi} - \cos(0) e^{0} = -e^{\pi} - 1 \]
Thus,
\[ \int_{0}^{\pi} e^{x} \cos x \, dx = -e^{\pi} - 1 + I \]

### Step 6: Substitute back into the equation
Now we substitute this back into our equation for \( I \):
\[ I = -(-e^{\pi} - 1 + I) \]
This simplifies to:
\[ I = e^{\pi} + 1 - I \]
Adding \( I \) to both sides gives:
\[ 2I = e^{\pi} + 1 \]
Thus, we find:
\[ I = \frac{e^{\pi} + 1}{2} \]

### Final Answer
The value of the integral is:
\[ \int_{0}^{\pi} e^{x} \sin x \, dx = \frac{e^{\pi} + 1}{2} \]