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
\(I=\frac{e^{2x}.sinx}{2}-\frac{1}{2}[cosx∫e^{2x}dx-∫(\frac{d}{dx}cosx)∫e^{2x}dx]dx\)
\(⇒I=\frac{e^{2x}sinx}{2}-\frac{1}{2}[cosx.\frac{e^{2x}}{2}-∫(-sinx)\frac{e^{2x}}{2}dx]\)
\(⇒I=\frac{e^{2x}sinx}{2}-\frac{1}{2}[\frac{e^{2x}cosx}{2}+\frac{1}{2}∫e^{2x}sinx dx]\)
\(⇒I=\frac{e^{2x}sinx}{2}-\frac{e^{2x}cosx}{4}-\frac{1}{4}I [From (1)]\)
\(⇒I=\frac{1}{4}I=\frac{e^{2x}.sinx}{2}-\frac{e^{2x}cosx}{4}\)
\(⇒\frac{5}{4}I=\frac{e^{2x}sinx}{2}-\frac{e^{2x}cosx}{4}\)
\(⇒I=\frac{4}{5}[\frac{e^{2x}sinx}{2}-\frac{e^{2x}cosx}{4}]+C\)
\(⇒I=\frac{e^{2x}}{5}[2sinx-cosx]+C\)