How Do You Solve the Integral of e^(2x)sin[3x] Using Integration by Parts?

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In summary, the integral of e^(2x)sin[3x] can be solved using integration by parts twice, or by using the elegant method of expressing sin(3x) as complex exponentials. Alternatively, one can also use tabular integration twice to obtain the same result.
  • #1
kennis2
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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 =(
 
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  • #2
What have you DONE? Show us what you have done and we may be able to point out your mistakes.
 
  • #3
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
I would consider that more difficult, but to each his own.
 
  • #6
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
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.]
 
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  • #8
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
Hippo,your integral is wrong.You differentiated the exponentials intead of integrating them...

Daniel.
 
  • #10
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.
 
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  • #11
[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
Yeah, I guess I could've expressed my answer in a better form.
 
  • #13
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 high school 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
[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: separate by parts until you get an equation that you can solve for the original integral.
 
  • #17
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 to How Do You Solve the Integral of e^(2x)sin[3x] Using Integration by Parts?

1. What is the formula for the integral of e^(2x)sin[3x]?

The formula for the integral of e^(2x)sin[3x] is ∫e^(2x)sin[3x]dx = (1/13)e^(2x)(3sin[3x]-2cos[3x]) + C.

2. How do you solve this integral using integration by parts?

To solve this integral using integration by parts, you would set u = sin[3x] and dv/dx = e^(2x). Then, integrate both sides to find v = (1/2)e^(2x), and use the formula ∫udv = uv - ∫vdu to solve for the integral.

3. Can you use trigonometric substitution to solve this integral?

Yes, you can use trigonometric substitution to solve this integral. You can substitute u = 3x and du = 3dx, which will change the integral to ∫e^(2/3)u*sin[u]du. Then, you can use the formula ∫e^(ax)sin[bx]dx = (1/a^2+b^2)e^(ax)(asin[bx]-bcos[bx]) + C to solve the integral.

4. Is this integral considered to be a difficult one to solve?

It depends on the individual's level of understanding and familiarity with integration techniques. For someone who is experienced and well-versed in integration, this integral may not be considered difficult. However, for someone who is new to integration, this integral may pose more of a challenge.

5. What are some real-life applications of this integral?

This integral has various applications in physics, engineering, and economics. For example, it can be used to model the growth of a population or the decay of radioactive substances. It can also be used to calculate the work done by a force acting on a moving object or the power output of an electric circuit. In economics, it can be used to calculate the present value of a future cash flow.

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