Question
Evaluate the following integrals: $\int\text{e}^{2\text{x}}\sin\text{x }\text{dx}$

Answer

Let $\text{I}=\int\text{e}^{2\text{x}}\sin\text{x }\text{dx}$
Considering $\sin x$ as first function and $e^{2x}$ as second function
$\text{I}=\sin\text{x}\frac{\text{e}^{2\text{x}}}{2}-\int\cos\text{x}\frac{\text{e}^{2\text{x}}}{2}\text{dx}$
$\Rightarrow\text{I}=\sin\text{x}\frac{\text{e}^{2\text{x}}}{2}-\frac{1}{2}\int\cos\text{e}^{2\text{x}}\text{dx}$
$\Rightarrow\text{I}=\frac{\sin\text{x }\text{e}^{2\text{x}}}{2}-\frac{1}{2}\Big[\cos\text{x}\frac{\text{e}^{2\text{x}}}{2}-\int(-\sin\text{x})\frac{\text{e}^{2\text{x}}}{2}\text{dx}\Big]$
$\Rightarrow\text{I}=\frac{\sin\text{x }\text{e}^{2\text{x}}}{2}-\frac{\cos\text{x }\text{e}^{2\text{x}}}{4}-\frac{1}{2}\int\frac{\text{e}^{2\text{x}}\sin\text{x}}{2}\text{dx}$
$\text{I}=\frac{\text{e}^{2\text{x}}(2\sin\text{x}-\cos\text{x})}{4}-\frac{\text{I}}{4}$
$\Rightarrow5\text{I}=\text{e}^{2\text{x}}(2\sin\text{x}-\cos\text{x})$
$\Rightarrow\text{I}=\frac{\text{e}^{2\text{x}}(2\sin\text{x}-\cos\text{x})}{5}+\text{C}$

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