Integral of e^(2x)sin[3x]?

  • Thread starter kennis2
  • Start date
  • #1
8
0
Integral of e^(2x)sin[3x]??

Integration by parts of: e^(2x)sin[3x]
result:e^2x(2sin3x-3cos3x)/13+C
can't get that result =(
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,833
961
What have you DONE? Show us what you have done and we may be able to point out your mistakes.
 
  • #3
789
4
Right, you do need to show your work. But I'll give you a hint. You're going to have to do integration by parts twice. Post what happens when you do this and you'll see how this problem can be manipulated to get the correct answer.
 
  • #4
Bah integration by parts is a long and tedious process... id express the sin 3x term as complex exponentials and work from there...

you would do well the recall that

[tex]sin({\omega}t)=\frac{e^{j{\omega}t}-e^{-j{\omega}t}}{2j}[/tex]
 
  • #5
789
4
I would consider that more difficult, but to each his own.
 
  • #6
dextercioby
Science Advisor
Homework Helper
Insights Author
13,016
566
Here's a very elegant way

[tex] \int e^{2x}\sin 3x \ dx=\mbox{Im}\left(\int e^{(2+3i)x} \ dx\right)=\mbox{Im}\left(\frac{e^{2x+3ix}}{2+3i}\right)+C=...=\frac{2\sin 3x-3\cos 3x}{13}e^{2x}+C [/tex]

1.I'll let u do the intermediate calculations...
2.Excercise:compute

[tex] \int e^{-x}\cos 4x \ dx [/tex]


Daniel.
 
  • #7
Hippo
Daniel, I can compute that in a really awkward way, expressing cos(4x) with complex exponentials.


[tex] \int e^{-x}\cos 4x \ dx = (\frac{1}{2}) \int e^{x(4j-1)} + e^{x(-4j-1)} \ dx[/tex]

[tex] = \frac{(-4j-1)e^{x(4j-1)} + (4j-1)e^{x(-4j-1)}}{34} + K[/tex]


How did you do the original integral, though?

[Edit: Yeah, I accidentally edited this instead of quoting it. I had to rewrite it.]
 
Last edited by a moderator:
  • #8
2,209
1
Its probably an example of integration by parts in his book, Jameson had the best hint, IMO as it was the one provided in my textbook.
 
  • #9
dextercioby
Science Advisor
Homework Helper
Insights Author
13,016
566
Hippo,your integral is wrong.You differentiated the exponentials intead of integrating them...

Daniel.
 
  • #10
Hippo
Look again.

[tex] \frac {e^{x(4j-1)}} {2(4j-1)} + \frac{e^{x(-4j-1)}}{2(-4j-1)} + K = \frac { (-4j-1) e^{x(4j-1)} + (4j-1) e^{x(-4j-1)} }{2*17} + K[/tex]


Now, will you please tell me how you did the original integral? I don't understand your method or notation.
 
Last edited by a moderator:
  • #11
dextercioby
Science Advisor
Homework Helper
Insights Author
13,016
566
[tex] \cos 4x=\frac{1}{2}\left(e^{4ix}+e^{-4ix}\right) [/tex]

[tex]e^{-x}\cos 4x=\frac{1}{2}\left[e^{x(4i-1)}+e^{x(-4i-1)}\right] [/tex]

[tex] \int e^{-x}\cos 4x \ dx=\frac{1}{2}\left[\int e^{x(4i-1)} \ dx+\int e^{x(-4i-1)} \ dx\right]=\frac{1}{2}\left[\frac{e^{x(4i-1)}}{4i-1}+\frac{e^{x(-4i-1)}}{-4i-1}\right]+C[/tex]

[tex]=\frac{1}{2}\left\{\frac{1}{17}\left[(-4i-1)e^{x(4i-1)}+(4i-1)e^{x(-4i-1)}\right]\right\}+C=\frac{1}{17}\left(4\sin 4x-\cos 4x\right)e^{-x} +C [/tex]

Daniel.
 
  • #12
Hippo
Yeah, I guess I could've expressed my answer in a better form. :grumpy:
 
  • #13
dextercioby
Science Advisor
Homework Helper
Insights Author
13,016
566
The algebra was pretty simple,just be careful with prducts of "i"-s and "-1"-s...


Daniel.
 
  • #14
dextercioby said:
Here's a very elegant way

[tex] \int e^{2x}\sin 3x \ dx=\mbox{Im}\left(\int e^{(2+3i)x} \ dx\right)=\mbox{Im}\left(\frac{e^{2x+3ix}}{2+3i}\right)+C=...=\frac{2\sin 3x-3\cos 3x}{13}e^{2x}+C [/tex]
Excuse my ignorance (I'm only a highschool student), but how did you get from

[tex] \int e^{2x}\sin 3x \ dx=\mbox{Im}\left(\int e^{(2+3i)x} \ dx\right) [/tex]

My knowledge of complex numbers is limited and I tried applying euler's formula/de moivre's theorem and got nowhere. Even working backwards doesn't seem to help. I was however able to do integration by parts without sweat.
 
  • #15
euler's formula
[tex]e^{ix}=cos(x)+isin(x)[/tex]

or some variation of that.
 
  • #16
68
0
[tex]\int e^{2x}sin(3x) dx = -\frac{1}{3}e^{2x}cos(3x) - \frac{2}{3}\int cos(3x) e^{2x} dx[/tex]
[tex]\int e^{2x}sin(3x) dx = -\frac{1}{3}e^{2x}cos(3x) - \frac{2}{3}(\frac{1}{3}e^{2x}sin(3x) - \frac{2}{3}\int e^{2x} sin(3x) dx) [/tex]
Solve for
[tex]\int e^{2x}sin(3x) dx[/tex]

Disclaimer: It's 12:46am here and I didn't bother to do this on paper, some negatives or fractions maybe wrong but the approach is correct: seperate by parts until you get an equation that you can solve for the original integral.
 
  • #17
36
0
why not just simply to tabular integration twice?

integral of e^2x * sin3x

original terms followed by deriv's/anti-deriv's

e^2x ------- sin3x
2e^2x ------ (-cos3x)/3
4e^2x ------ (-sin3x)/9


which then yields

=-(cos3x * e^2x)/3 + (2/9)(e^2x * sin3x) - (4/9) ( integral of e^2x * sin3x)


now take that last integral, add it to both sides , thus leaving



(9/13)[-(cos3x * e^2x)/3 + (2/9)(e^2x * sin3x)] + C
 

Related Threads on Integral of e^(2x)sin[3x]?

Replies
4
Views
698
  • Last Post
Replies
2
Views
14K
  • Last Post
Replies
5
Views
5K
  • Last Post
Replies
3
Views
9K
  • Last Post
Replies
13
Views
8K
Replies
5
Views
6K
  • Last Post
Replies
17
Views
14K
  • Last Post
Replies
8
Views
25K
  • Last Post
Replies
3
Views
4K
  • Last Post
Replies
4
Views
12K
Top