- #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 =(

- Thread starter kennis2
- Start date

- #1

- 8

- 0

Integration by parts of: e^(2x)sin[3x]

result:e^2x(2sin3x-3cos3x)/13+C

can't get that result =(

- #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

- #4

- 259

- 0

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

- 13,016

- 566

[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.]

[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

- #9

- 13,016

- 566

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.

[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

- 13,016

- 566

[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

- 13,016

- 566

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

Daniel.

Daniel.

- #14

- 17

- 0

Excuse my ignorance (I'm only a highschool student), but how did you get fromdextercioby 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]

[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

- 94

- 0

euler's formula

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

or some variation of that.

[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}(\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

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

- 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