Question
Find:
$\int \text{e}^{2\text{x}} \sin \text{(3x + 1)} \text{ dx}$

Answer

$\text{I} = \int \text{e}^{2\text{x}} \sin \text{(3x + 1)} \text{ dx}$
$= \sin \text{(3x + 1)}. \frac{\text{e}^{2x}}{2} - \int 3 \cos \text{(3x + 1)} . \frac{\text{e}^{2\text{x}}}{2} \text{dx}$
$= \frac{\text{e}^{\text{2x}}}{2}. \sin \text{(3x + 1)} - \frac{3}{2} \bigg[\cos \text{(3x + 1)} . \frac{\text{e}^{\text{2x}}}{2} - \int -3 \sin \text{(3x + 1)}. \frac{\text{e}^{\text{2x}}}{2} \text{dx}\bigg]$
$= \frac{\text{e}^{\text{2x}}}{2} \sin \text{(3x + 1)} - \frac{3}{4} \cos \text{(3x + 1)} . \text{e}^{\text{2x}} - \frac{9}{4} \text{I + c}$
$\Rightarrow \frac{13}{4} \text{I} = \frac{\text{e}^{\text{2x}}}{4} [2 \sin \text{(3x + 1)} - 3 \cos \text{(3x + 1)}] + \text{c}$
$\Rightarrow \text{I} = \frac{\text{e}^{\text{2x}}}{13} [ 2 \sin \text{(3x + 1)} - 3 \cos \text{(3x + 1)}] + \text{c}$

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