$\int e^{2x}(\sin x + \frac{1}{2}\cos x) dx$ is equal to

When integrating $e^{ax}$ multiplied with a combination of $\sin x$ and $\cos x$, the result typically has the form $e^{ax}$ times another trigonometric combination.

For this integral, consider the function $\frac{1}{2}e^{2x}\sin x$.

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To check if this is correct, differentiate $\frac{1}{2}e^{2x}\sin x$ using the product rule:

$\frac{d}{dx}\left[\frac{1}{2}e^{2x}\sin x\right]$

$= \frac{1}{2}\left[(e^{2x})' \cdot \sin x + e^{2x} \cdot (\sin x)'\right]$

$= \frac{1}{2}\left[2e^{2x} \cdot \sin x + e^{2x} \cdot \cos x\right]$

$= e^{2x}\sin x + \frac{1}{2}e^{2x}\cos x$

$= e^{2x}\left(\sin x + \frac{1}{2}\cos x\right)$

This matches the integrand exactly.

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Therefore:

$\int e^{2x}\left(\sin x + \frac{1}{2}\cos x\right) dx = \frac{1}{2}e^{2x}\sin x + C$