Integrate the function: \(e^{2x} sinx\)
Solution and Explanation
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
\(I=\frac{e^{2x}.sinx}{2}-\frac{1}{2}[cosx∫e^{2x}dx-∫(\frac{d}{dx}cosx)∫e^{2x}dx]dx\)
\(⇒I=\frac{e^{2x}sinx}{2}-\frac{1}{2}[cosx.\frac{e^{2x}}{2}-∫(-sinx)\frac{e^{2x}}{2}dx]\)
\(⇒I=\frac{e^{2x}sinx}{2}-\frac{1}{2}[\frac{e^{2x}cosx}{2}+\frac{1}{2}∫e^{2x}sinx dx]\)
\(⇒I=\frac{e^{2x}sinx}{2}-\frac{e^{2x}cosx}{4}-\frac{1}{4}I [From (1)]\)
\(⇒I=\frac{1}{4}I=\frac{e^{2x}.sinx}{2}-\frac{e^{2x}cosx}{4}\)
\(⇒\frac{5}{4}I=\frac{e^{2x}sinx}{2}-\frac{e^{2x}cosx}{4}\)
\(⇒I=\frac{4}{5}[\frac{e^{2x}sinx}{2}-\frac{e^{2x}cosx}{4}]+C\)
\(⇒I=\frac{e^{2x}}{5}[2sinx-cosx]+C\)
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Concepts Used:
Integration by Partial Fractions
The number of formulas used to decompose the given improper rational functions is given below. By using the given expressions, we can quickly write the integrand as a sum of proper rational functions.
For examples,
