`int e^(x). " sin x( sin x+ 2 cos x) dx "`

Answer

To evaluate the integral \( I = \int e^x \sin x (\sin x + 2 \cos x) \, dx \), we can follow these steps:

### Step 1: Rewrite the Integral
First, we can rewrite the integral as:
\[ I = \int e^x \left( \sin^2 x + 2 \sin x \cos x \right) \, dx \]

### Step 2: Use a Trigonometric Identity
Recall that \( \sin^2 x = \frac{1 - \cos(2x)}{2} \) and \( 2 \sin x \cos x = \sin(2x) \). Therefore, we can rewrite the integral as:
\[ I = \int e^x \left( \frac{1 - \cos(2x)}{2} + \sin(2x) \right) \, dx \]
This simplifies to:
\[ I = \int e^x \left( \frac{1}{2} - \frac{1}{2} \cos(2x) + \sin(2x) \right) \, dx \]

### Step 3: Split the Integral
Now we can split the integral into three parts:
\[ I = \frac{1}{2} \int e^x \, dx - \frac{1}{2} \int e^x \cos(2x) \, dx + \int e^x \sin(2x) \, dx \]

### Step 4: Evaluate Each Integral
1. The first integral:
\[ \frac{1}{2} \int e^x \, dx = \frac{1}{2} e^x + C_1 \]

2. For the second integral \( \int e^x \cos(2x) \, dx \), we can use integration by parts or a known result:
\[ \int e^x \cos(2x) \, dx = \frac{e^x}{5} (2 \cos(2x) + \sin(2x)) + C_2 \]

3. For the third integral \( \int e^x \sin(2x) \, dx \), we again use integration by parts or a known result:
\[ \int e^x \sin(2x) \, dx = \frac{e^x}{5} (\sin(2x) - 2 \cos(2x)) + C_3 \]

### Step 5: Combine the Results
Now we combine all the results:
\[ I = \frac{1}{2} e^x - \frac{1}{2} \left( \frac{e^x}{5} (2 \cos(2x) + \sin(2x)) \right) + \left( \frac{e^x}{5} (\sin(2x) - 2 \cos(2x)) \right) + C \]

### Step 6: Simplify the Expression
Combining the terms will give us the final result:
\[ I = \frac{1}{2} e^x - \frac{1}{10} e^x (2 \cos(2x) + \sin(2x)) + \frac{1}{5} e^x (\sin(2x) - 2 \cos(2x)) + C \]

### Final Answer
Thus, the final answer is:
\[ I = e^x \left( \frac{1}{2} - \frac{1}{10} (2 \cos(2x) + \sin(2x)) + \frac{1}{5} (\sin(2x) - 2 \cos(2x)) \right) + C \]