Read the following text and answer the followig questions on the basis of the same :
`inte^(x)[f(x) + f'(x)]dx = int e^(x)f(x)dx + int e^(x) f'(x)dx`
`= f(x)e^(x) - int f'(x)e^(x)dx + int f'(x)e^(x)dx`
`= e^(x)f(x) + c`
`int e^(x)(sin x + cos x)dx` =

Options

A. `e^(x)cos x + c`
B. `e^(x) sin x + c`
C. `e^(x) + c`
D. `e^(x)(-cos x + sin x ) + c`

Answer

To solve the integral \( \int e^{x}(\sin x + \cos x)dx \), we will use the formula derived from the integration of the product of an exponential function and a function along with its derivative.

### Step-by-Step Solution:

1. **Identify \( f(x) \) and \( f'(x) \)**:
- Here, we can let \( f(x) = \sin x \).
- Therefore, the derivative \( f'(x) = \cos x \).

2. **Rewrite the integral**:
- The integral can be rewritten as:
\[ \int e^{x}(\sin x + \cos x)dx = \int e^{x}\sin x \, dx + \int e^{x}\cos x \, dx \]

3. **Use the integration formula**:
- According to the integration formula:
\[ \int e^{x}[f(x) + f'(x)]dx = e^{x}f(x) + C \]
- Applying this to our case:
\[ \int e^{x}(\sin x + \cos x)dx = e^{x}\sin x + C \]

4. **Final Result**:
- Thus, the integral evaluates to:
\[ \int e^{x}(\sin x + \cos x)dx = e^{x}\sin x + C \]

### Final Answer:
\[ \int e^{x}(\sin x + \cos x)dx = e^{x}\sin x + C \]

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