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

What have you DONE? Show us what you have done and we may be able to point out your mistakes.

 

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.

 

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]

 

I would consider that more difficult, but to each his own.

 

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.

 

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.]

 

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.

 

Hippo,your integral is wrong.You differentiated the exponentials intead of integrating them...

Daniel.

 

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.

 

[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.

 

Yeah, I guess I could've expressed my answer in a better form. :grumpy:

 

The algebra was pretty simple,just be careful with prducts of "i"-s and "-1"-s...


Daniel.

 

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.

 

euler's formula
[tex]e^{ix}=cos(x)+isin(x)[/tex]

or some variation of that.

 

[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.

 

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