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

[kennis2]

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 =(

 

[HallsofIvy]

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

 

[Jameson]

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.

 

[Theelectricchild]

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

$$sin({\omega}t)=\frac{e^{j{\omega}t}-e^{-j{\omega}t}}{2j}$$

 

[Jameson]

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

 

[dextercioby]

Here's a very elegant way

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

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

$$\int e^{-x}\cos 4x \ dx$$


Daniel.

 

[Hippo]

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


$$\int e^{-x}\cos 4x \ dx = (\frac{1}{2}) \int e^{x(4j-1)} + e^{x(-4j-1)} \ dx$$

$$= \frac{(-4j-1)e^{x(4j-1)} + (4j-1)e^{x(-4j-1)}}{34} + K$$


How did you do the original integral, though?

[Edit: Yeah, I accidentally edited this instead of quoting it. I had to rewrite it.][/color]

 

[whozum]

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.

 

[dextercioby]

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

Daniel.

 

[Hippo]

Look again.

$$\frac {e^{x(4j-1)}} {2(4j-1)} + \frac{e^{x(-4j-1)}}{2(-4j-1)} + K = \frac { (-4j-1) e^{x(4j-1)} + (4j-1) e^{x(-4j-1)} }{2*17} + K$$


Now, will you please tell me how you did the original integral? I don't understand your method or notation.

 

[dextercioby]

$$\cos 4x=\frac{1}{2}\left(e^{4ix}+e^{-4ix}\right)$$

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

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

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

Daniel.

 

[Hippo]

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

 

[dextercioby]

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


Daniel.

 

[SplinterIon]

Here's a very elegant way

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

Excuse my ignorance (I'm only a high school student), but how did you get from

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

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.

 

[p53ud0 dr34m5]

euler's formula
$$e^{ix}=cos(x)+isin(x)$$

or some variation of that.

 

[µ³]

$$\int e^{2x}sin(3x) dx = -\frac{1}{3}e^{2x}cos(3x) - \frac{2}{3}\int cos(3x) e^{2x} dx$$
$$\int e^{2x}sin(3x) dx = -\frac{1}{3}e^{2x}cos(3x) - \frac{2}{3}(\frac{1}{3}e^{2x}sin(3x) - \frac{2}{3}\int e^{2x} sin(3x) dx)$$
Solve for
$$\int e^{2x}sin(3x) dx$$

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.

 

[Hessam]

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