Integration by parts

In this instance it doesn't matter much, it's probably easier to use the trig term.

You will require two iterations, ending up with a similar integral to the one above when you try and work it out the second time.

Calling the integral I, you can then rewrite in the form I = ...

Reply 2

Aristotle's' Disciple

19

Call du/dx e^3x. Easier to intergrate. sin2x can be differentiated easily enough. Then just apply the rule.

Reply 3

boromir9111

Not sure if this will help but kinda gives an explanation!

http://www.wolframalpha.com/input/?i=integrate+e^%283x%29*sin%282x%29

Reply 4

OP

Original post by marcusmerehay

In this instance it doesn't matter much, it's probably easier to use the trig term.

You will require two iterations, ending up with a similar integral to the one above when you try and work it out the second time.

Calling the integral I, you can then rewrite in the form I = ...

oh so I'll still have to do in in two stages then?

Reply 5

Original post by Plato's Trousers

given the integral

\displaystyle\int e^{3x}\sin(2x)\;dx

presumably you use integration by parts, but which bit do I call v and which bit du/dx ?

Actually in general, what's the strategy for deciding which bit to call what? I thought you choose the bit that differentiates to a constant to be v, but in this case, neither of them do.

:confused:

Ta

Alternatively, if you want to avoid IBP you could notice that:

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

,

and find that imaginary part.

(edited 13 years ago)

Reply 6

OP

Original post by boromir9111

Not sure if this will help but kinda gives an explanation!

http://www.wolframalpha.com/input/?i=integrate+e^%283x%29*sin%282x%29

yeah, I tried Walpha, but it doesn't do it by parts. It uses some kind of general formula

Reply 7

OP

Original post by Farhan.Hanif93

Alternatively, you could notice that:

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

,

and find that imaginary part.

um, you're joking right? (you know I won't understand that! :angry:

)

Reply 8

Potassium^2

Original post by Plato's Trousers

Actually in general, what's the strategy for deciding which bit to call what?

Try googling LIATE Rule, its a rule of thumb which I follow for integration by parts.
Basically you choose 'u' by what come first in the list: Logs, Inverse trig, Algebra, Trig, Exponentials

Reply 9

Original post by Plato's Trousers

um, you're joking right? (you know I won't understand that! :angry:

)

My apologies. I thought you were doing an Open University course on it so assumed that you had met complex numbers properly by now. :o:

Do you follow what the others have suggested?

Reply 10

eliotball

7

http://en.wikipedia.org/wiki/Integration_by_parts

In Examples, from 'An unusual example commonly used...'

Reply 11

OP

Original post by Potassium^2

Try googling LIATE Rule, its a rule of thumb which I follow for integration by parts.
Basically you choose 'u' by what come first in the list: Logs, Inverse trig, Algebra, Trig, Exponentials

that's cool! I haven't heard of that before.

Original post by Farhan.Hanif93

My apologies. I thought you were doing an Open University course on it so assumed that you had met complex numbers properly by now. :o:

Do you follow what the others have suggested?

well, I know what complex numbers are, but I don't see the relevance to this.

Doing it by parts just keeps on generating new products? So I don't see how it helps.

Choosing dv/dx to be e^3x I get

Unparseable latex formula:

\dfrac{1}{3}3e^{3x}\sin(2x)}-\displaystyle\int\dfrac{2}{3}e^{3x}\cos(2x)\;dx

which doesn't really help me, as I have still got a product to integrate. And you keep on getting a product, presumably?

(edited 13 years ago)

Reply 12

Original post by Plato's Trousers

well, I know what complex numbers are, but I don't see the relevance to this.

I take it that you've never seen De Moivre's theorem before? Or the definition:

e^{i\theta} = \cos \theta +i\sin \theta

?

Doing it by parts just keeps on generating new products? So I don't see how it helps.

Choosing dv/dx to be e^3x I get

\dfrac{e^{3x}\sin(2x)}{3}-\displaystyle\int\dfrac{2}{3}e^{3x}\cos(2x)\;dx

which doesn't really help me, as I have still got a product to integrate. And you keep on getting a product, presumably?

Use IBP again with

\frac{dv}{dx}=e^{3x}

and notice the integral part that emerges. It will be similar to what you started with. Therefore, if you let

I=\displaystyle\int e^{3x}\sin (2x)dx

, you will have something of the form:

I=[\mathrm{stuff}] + kI

where k is a constant.

Reply 13

OP

Original post by Farhan.Hanif93

Alternatively, if you want to avoid IBP you could notice that:

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

,

and find that imaginary part.

oh wait, yes, sorry! I see what you mean now. You're using Euler's relation. That's sneaky :ninja:

It's just that you first had a Hebrew letter in place of the Im, (before you edited it). Aleph, maybe? I didn't know what that was

Reply 14

Original post by Plato's Trousers

oh wait, yes, sorry! I see what you mean now. You're using Euler's relation. That's sneaky :ninja:

It's just that you first had a Hebrew letter in place of the Im, (before you edited it). Aleph, maybe? I didn't know what that was

Good to hear that you've got the idea.

Yeah, I don't know what happened there, slip of the finger I think.

Reply 15

OP

Original post by Farhan.Hanif93

Good to hear that you've got the idea.

Yeah, I don't know what happened there, slip of the finger I think.

our posts crossed I think. You were posting about de Moivre and I was posting about Euler. I always get those mixed up, what's the difference?

Reply 16

AgeOfAquarius

Original post by Farhan.Hanif93

I=[\mathrm{stuff}] + kI

where k is a constant.

Which won't work if k = 1.

Reply 17