Question

Integrate the function $e^{2x}\sin x$

Answer 1

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