Step 1 : Understanding the Question:
This problem asks us to find the integral of an exponential function multiplied by the sum of sine and cosine functions. While this could be solved using integration by parts twice, it is specifically designed to test the recognition of a standard integral form. Recognizing this pattern allows for a much faster and less error-prone solution path.
Step 2 : Key Formulas and approach:
We utilize the special integral formula:
\[ \int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C \]
This identity is derived from the product rule of differentiation in reverse. If we differentiate $e^x f(x)$, we get $e^x f(x) + e^x f'(x)$. Therefore, the integral of $e^x(f(x) + f'(x))$ must be $e^x f(x)$. Our approach is to identify a function $f(x)$ in the integrand such that its derivative $f'(x)$ is also present in the sum.
Step 3 : Detailed Explanation:
Let the integrand be $I = \int e^x (\sin x + \cos x) dx$.
We compare this to the form $e^x [f(x) + f'(x)]$.
Let $f(x) = \sin x$.
Now, calculate the derivative: $f'(x) = \frac{d}{dx}(\sin x) = \cos x$.
The given expression perfectly matches this form, as $(\sin x + \cos x)$ is exactly $[f(x) + f'(x)]$.
Applying the theorem directly: $\int e^x (\sin x + \cos x) dx = e^x (\sin x) + C$.
As a verification, if we differentiate $e^x \sin x$, we get $(e^x \cdot \sin x) + (e^x \cdot \cos x)$ by the product rule, which confirms our result.
Step 4 : Final Answer:
The integral is $e^x \sin x + C$, which corresponds to option (B).