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    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
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    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 = ...
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    Call du/dx e^3x. Easier to intergrate. sin2x can be differentiated easily enough. Then just apply the rule.
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    Not sure if this will help but kinda gives an explanation!

    http://www.wolframalpha.com/input/?i...29*sin%282x%29
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    (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?
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    (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.
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    (Original post by boromir9111)
    Not sure if this will help but kinda gives an explanation!

    http://www.wolframalpha.com/input/?i...29*sin%282x%29
    yeah, I tried Walpha, but it doesn't do it by parts. It uses some kind of general formula
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    (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:)
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    (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
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    (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?
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    http://en.wikipedia.org/wiki/Integration_by_parts

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

    \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?
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    (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.
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    (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
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    (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.
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    (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?
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    (Original post by Farhan.Hanif93)
    I=[\mathrm{stuff}] + kI where k is a constant.
    Which won't work if k = 1.
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    (Original post by AgeOfAquarius)
    Which won't work if k = 1.
    What's your point? I haven't indicated that k is 1. k isn't arbitrary and I have used an equality sign as opposed to an equivalence sign for a reason!
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    (Original post by Plato's Trousers)
    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?
    A quick google will tell you the answer to your question.
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    (Original post by Farhan.Hanif93)
    What's your point? I haven't indicated that k is 1.
    You misunderstood me. It was just a caution for the OP.
 
 
 
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