$e^{2 x} \sin x$
Let $I=\int e^{2 x} \sin x d x$ ...(1)
Integrating by parts, we obtain
$I=\sin x \int e^{2 x} d x-\int\left\{\left(\frac{d}{d x} \sin x\right) \int e^{2 x} d x\right\} d x$
$\Rightarrow I=\sin x \cdot \frac{e^{2 x}}{2}-\int \cos x \cdot \frac{e^{2 x}}{2} d x$
$\Rightarrow I=\frac{e^{2 x} \sin x}{2}-\frac{1}{2} \int e^{2 x} \cos x d x$
Again integrating by parts, we obtain
$I=\frac{e^{2 x} \cdot \sin x}{2}-\frac{1}{2}\left[\cos x \int e^{2 x} d x-\int\left\{\left(\frac{d}{d x} \cos x\right) \int e^{2 x} d x\right\} d x\right.$
$\Rightarrow I=\frac{e^{2 x} \sin x}{2}-\frac{1}{2}\left[\cos x \cdot \frac{e^{2 x}}{2}-\int(-\sin x) \frac{e^{2 x}}{2} d x\right]$
$\Rightarrow I=\frac{e^{2 x} \cdot \sin x}{2}-\frac{1}{2}\left[\frac{e^{2 x} \cos x}{2}+\frac{1}{2} \int e^{2 x} \sin x d x\right]$
$\Rightarrow I=\frac{e^{2 x} \sin x}{2}-\frac{e^{2 x} \cos x}{4}-\frac{1}{4} I$ [From (1)]
$\Rightarrow I+\frac{1}{4} I=\frac{e^{2 x} \cdot \sin x}{2}-\frac{e^{2 x} \cos x}{4}$
$\Rightarrow \frac{5}{4} I=\frac{e^{2 x} \sin x}{2}-\frac{e^{2 x} \cos x}{4}$
$\Rightarrow I=\frac{4}{5}\left[\frac{e^{2 x} \sin x}{2}-\frac{e^{2 x} \cos x}{4}\right]+\mathrm{C}$
$\Rightarrow I=\frac{e^{2 x}}{5}[2 \sin x-\cos x]+\mathrm{C}$
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