`int_(0)^(pi) e^(x) sin 2 x dx=`

Options

A. `2/5 ( 1-e^(pi))`
B. `2/5 (e^(pi-1))`
C. `1/5 (1-e^(pi))`
D. `1/5 (e^(pi)-1)`

Answer

To solve the integral \( I = \int_{0}^{\pi} e^x \sin(2x) \, dx \), we will use integration by parts.

### Step 1: Set up the integration by parts
We will use the formula for integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
Let:
- \( u = \sin(2x) \) (so \( du = 2 \cos(2x) \, dx \))
- \( dv = e^x \, dx \) (so \( v = e^x \))

### Step 2: Apply integration by parts
Using the integration by parts formula:
\[ I = \left[ \sin(2x) e^x \right]_{0}^{\pi} - \int_{0}^{\pi} e^x (2 \cos(2x)) \, dx \]
Calculating the first term:
\[ \left[ \sin(2x) e^x \right]_{0}^{\pi} = \sin(2\pi) e^{\pi} - \sin(0) e^{0} = 0 - 0 = 0 \]
Thus, we have:
\[ I = -2 \int_{0}^{\pi} e^x \cos(2x) \, dx \]

### Step 3: Solve the new integral
Let \( J = \int_{0}^{\pi} e^x \cos(2x) \, dx \). We will again use integration by parts:
Let:
- \( u = \cos(2x) \) (so \( du = -2 \sin(2x) \, dx \))
- \( dv = e^x \, dx \) (so \( v = e^x \))

Now applying integration by parts:
\[ J = \left[ \cos(2x) e^x \right]_{0}^{\pi} - \int_{0}^{\pi} e^x (-2 \sin(2x)) \, dx \]
Calculating the first term:
\[ \left[ \cos(2x) e^x \right]_{0}^{\pi} = \cos(2\pi) e^{\pi} - \cos(0) e^{0} = 1 \cdot e^{\pi} - 1 \cdot 1 = e^{\pi} - 1 \]
Thus, we have:
\[ J = (e^{\pi} - 1) + 2 \int_{0}^{\pi} e^x \sin(2x) \, dx = (e^{\pi} - 1) + 2I \]

### Step 4: Substitute back into the equation
Now we have:
\[ I = -2J \]
Substituting for \( J \):
\[ I = -2 \left( (e^{\pi} - 1) + 2I \right) \]
Expanding this gives:
\[ I = -2(e^{\pi} - 1) - 4I \]
Rearranging terms:
\[ I + 4I = -2(e^{\pi} - 1) \]
\[ 5I = -2(e^{\pi} - 1) \]
Thus:
\[ I = \frac{-2(e^{\pi} - 1)}{5} \]
This simplifies to:
\[ I = \frac{2 - 2e^{\pi}}{5} \]

### Final Answer
\[ \int_{0}^{\pi} e^x \sin(2x) \, dx = \frac{2 - 2e^{\pi}}{5} \]