What is `int_0^pi e^x sin xdx` equal to

Options

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

Answer

To solve the integral \( I = \int_0^\pi e^x \sin x \, dx \), we will use integration by parts.

### Step-by-Step Solution:

1. **Identify \( u \) and \( dv \)**:
We choose:
\[ u = e^x \quad \text{and} \quad dv = \sin x \, dx \]
Then, we differentiate and integrate to find \( du \) and \( v \):
\[ du = e^x \, dx \quad \text{and} \quad v = -\cos x \]

2. **Apply Integration by Parts**:
Using the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
We substitute our values:
\[ I = \left[ e^x (-\cos x) \right]_0^\pi - \int_0^\pi (-\cos x) e^x \, dx \]
Simplifying this gives:
\[ I = -e^x \cos x \bigg|_0^\pi + \int_0^\pi e^x \cos x \, dx \]

3. **Evaluate the Boundary Terms**:
Calculate \( -e^x \cos x \) at the limits:
\[ -e^\pi \cos(\pi) + e^0 \cos(0) = -e^\pi (-1) + 1 = e^\pi + 1 \]

4. **Substitute Back into the Integral**:
Now we have:
\[ I = e^\pi + 1 + \int_0^\pi e^x \cos x \, dx \]

5. **Integrate \( \int_0^\pi e^x \cos x \, dx \)**:
We will again use integration by parts for this integral. Let:
\[ u = e^x \quad \text{and} \quad dv = \cos x \, dx \]
Then:
\[ du = e^x \, dx \quad \text{and} \quad v = \sin x \]
Applying integration by parts:
\[ \int e^x \cos x \, dx = e^x \sin x \bigg|_0^\pi - \int_0^\pi \sin x \, e^x \, dx \]

6. **Evaluate the Boundary Terms Again**:
Calculate \( e^x \sin x \) at the limits:
\[ e^\pi \sin(\pi) - e^0 \sin(0) = e^\pi (0) - 0 = 0 \]
Thus:
\[ \int_0^\pi e^x \cos x \, dx = 0 - \int_0^\pi e^x \sin x \, dx = -I \]

7. **Combine Results**:
Substitute back:
\[ I = e^\pi + 1 - I \]
Rearranging gives:
\[ 2I = e^\pi + 1 \]
Therefore:
\[ I = \frac{e^\pi + 1}{2} \]

### Final Answer:
\[ \int_0^\pi e^x \sin x \, dx = \frac{e^\pi + 1}{2} \]