Integrate the function in Exercise: e^ 2x

Question

Integrate the function in Exercise:
$\text{e}^{2\text{x}}\sin\text{x}$

✓

Answer

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