Question

\( \int e^x (\sin x + 2 \cos x) \sin x \, dx \) is equal to:

• A. \( e^x \cos x + C \)
• B. \( e^x \sin x + C \)
• C. \( e^x \sin^2 x + C \)
• D. \( e^x \sin 2x + C \)
• E. \( e^x (\cos x + \sin x) + C \)

Hint:

The derivative of $\sin^2 x$ is $\sin 2x$. Recognizing this relationship immediately solves many $e^x$ integration problems involving squares of trigonometric functions.

Answer 1

The Correct Option is C

Concept: Like the previous problem, we use the identity \( \int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C \). We must distribute the \( \sin x \) term first.

Step 1: Distribute the terms.
\[ \int e^x (\sin^2 x + 2 \sin x \cos x) \, dx \]

Step 2: Identify \( f(x) \) and \( f'(x) \).
Let \( f(x) = \sin^2 x \). Using the chain rule: \[ f'(x) = \frac{d}{dx}(\sin x)^2 = 2 \sin x \cdot \frac{d}{dx}(\sin x) = 2 \sin x \cos x \] Note that this is also equal to \( \sin 2x \), but the form \( 2 \sin x \cos x \) is explicitly present in our integrand.

Step 3: Apply the property.
The integral is in the form \( \int e^x [f(x) + f'(x)] \, dx \): \[ \text{Integral} = e^x \sin^2 x + C \]