Question

Integrate the function: $e^{2x} sinx$

Answer 1

The correct answer is: $I=\frac{e^{2x}}{5}[2sinx-cosx]+C$
Let $I=∫e^{2x} sinx dx...(1)$
Integrating by parts,we obtain
$I=sinx∫e^{2x} dx-∫[(\frac{d}{dx}-sinx)∫e^{2x} dx]dx$
$⇒I=sinx.\frac{e^{2x}}{2}-∫cosx.\frac{e^{2x}}{2}dx$
$⇒I=e^{2x}sin\frac{x}{2}-\frac{1}{2}∫e^{2x}cosx\, dx$
Again integrating by parts,we obtain
$I=\frac{e^{2x}.sinx}{2}-\frac{1}{2}[cosx∫e^{2x}dx-∫(\frac{d}{dx}cosx)∫e^{2x}dx]dx$
$⇒I=\frac{e^{2x}sinx}{2}-\frac{1}{2}[cosx.\frac{e^{2x}}{2}-∫(-sinx)\frac{e^{2x}}{2}dx]$
$⇒I=\frac{e^{2x}sinx}{2}-\frac{1}{2}[\frac{e^{2x}cosx}{2}+\frac{1}{2}∫e^{2x}sinx dx]$
$⇒I=\frac{e^{2x}sinx}{2}-\frac{e^{2x}cosx}{4}-\frac{1}{4}I [From (1)]$
$⇒I=\frac{1}{4}I=\frac{e^{2x}.sinx}{2}-\frac{e^{2x}cosx}{4}$
$⇒\frac{5}{4}I=\frac{e^{2x}sinx}{2}-\frac{e^{2x}cosx}{4}$
$⇒I=\frac{4}{5}[\frac{e^{2x}sinx}{2}-\frac{e^{2x}cosx}{4}]+C$
$⇒I=\frac{e^{2x}}{5}[2sinx-cosx]+C$