Integrate: `int e^x sin^2 x dx.`

Answer

To solve the integral \( \int e^x \sin^2 x \, dx \), we can follow these steps:

### Step 1: Rewrite \( \sin^2 x \)
Using the identity for sine squared, we have:
\[ \sin^2 x = \frac{1 - \cos(2x)}{2} \]
Thus, we can rewrite the integral as:
\[ \int e^x \sin^2 x \, dx = \int e^x \cdot \frac{1 - \cos(2x)}{2} \, dx \]
This simplifies to:
\[ \frac{1}{2} \int e^x (1 - \cos(2x)) \, dx \]

### Step 2: Split the Integral
Now, we can split the integral into two parts:
\[ \frac{1}{2} \left( \int e^x \, dx - \int e^x \cos(2x) \, dx \right) \]

### Step 3: Integrate \( \int e^x \, dx \)
The integral of \( e^x \) is straightforward:
\[ \int e^x \, dx = e^x + C_1 \]

### Step 4: Integrate \( \int e^x \cos(2x) \, dx \)
To integrate \( \int e^x \cos(2x) \, dx \), we can use integration by parts or a known formula. The formula for integrating \( e^{ax} \cos(bx) \) is:
\[ \int e^{ax} \cos(bx) \, dx = \frac{e^{ax}}{a^2 + b^2} (a \cos(bx) + b \sin(bx)) + C \]
Here, \( a = 1 \) and \( b = 2 \):
\[ \int e^x \cos(2x) \, dx = \frac{e^x}{1^2 + 2^2} (1 \cdot \cos(2x) + 2 \cdot \sin(2x)) = \frac{e^x}{5} (\cos(2x) + 2 \sin(2x)) + C_2 \]

### Step 5: Combine the Results
Now, substituting back into our integral:
\[ \int e^x \sin^2 x \, dx = \frac{1}{2} \left( e^x - \frac{e^x}{5} (\cos(2x) + 2 \sin(2x)) \right) + C \]
This simplifies to:
\[ \int e^x \sin^2 x \, dx = \frac{1}{2} e^x - \frac{1}{10} e^x (\cos(2x) + 2 \sin(2x)) + C \]

### Final Answer
Thus, the final result is:
\[ \int e^x \sin^2 x \, dx = \frac{1}{2} e^x - \frac{1}{10} e^x (\cos(2x) + 2 \sin(2x)) + C \]