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    (Original post by AgeOfAquarius)
    You misunderstood me. It was a just a caution for the OP.
    I don't understand how k=1 can even arise by accident.
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    (Original post by RK92)
    ...
    Any reason for negging my post that actually helped the OP?
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    (Original post by Plato's Trousers)
    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?
    First and foremost-use the LIATE/ILATE rule to decide which is u and which is v.

    Then-

    I=int. e^3x sin 2x dx
    =sin 2x int e^3x dx - int ( d(sin 2x)/dx . int e^3x dx ) dx
    = 1/3 e^3x sin 2x - int( d(sin 2x)/dx . int e^3x dx)) dx
    = 1/3 e^3x sin 2x - int(2 cos 2x . 1/3 e^3x) dx
    = 1/3 e^3x sin 2x - 2/3 int(cos 2x e^3x) dx
    = 1/3 e^3x sin 2x - 2/3 [ 1/3 e^3x cos 2x - int { d(cos 2x)/dx . int (e^3x) dx} dx ]
    = 1/3 e^3x sin 2x - 2/3 [ 1/3 e^3x cos 2x - int { (-2 sin 2x) . 1/3 e^3x}dx ]
    = 1/3 e^3x sin 2x - 2/3 [1/3 e^3x cos 2x + 2/3 int(sin 2x e^3x) dx]
    = 1/3 e^3x sin 2x - 2/9 e^3x cos 2x - 4/9 I

    Therefore,
    13/9 I = 1/3 e^3x sin 2x - 2/9 e^3x cos 2x

    I= 9/13 [1/3 e^3x sin 2x - 2/9 e^3x cos 2x] + k


    ..typing that was tough, so I'm sorry if I made any typing errors, but this is the sum.


    EDIT: I just typed that whole thing out on my cell phone - how can someone neg me?!
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    (Original post by Farhan.Hanif93)
    I don't understand how k=1 can even arise by accident.
    Oh, trust me; it does.
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    (Original post by aankhi)
    First and foremost-use the LIATE/ILATE rule to decide which is u and which is v.

    Then-

    I=int. e^3x sin 2x dx
    =sin 2x int e^3x dx - int ( d(sin 2x)/dx . int e^3x dx ) dx
    = 1/3 e^3x sin 2x - int( d(sin 2x)/dx . int e^3x dx)) dx
    = 1/3 e^3x sin 2x - int(2 cos 2x . 1/3 e^3x) dx
    = 1/3 e^3x sin 2x - 2/3 int(cos 2x e^3x) dx
    = 1/3 e^3x sin 2x - 2/3 [ 1/3 e^3x cos 2x - int { d(cos 2x)/dx . int (e^3x) dx} dx ]
    = 1/3 e^3x sin 2x - 2/3 [ 1/3 e^3x cos 2x - int { (-2 sin 2x) . 1/3 e^3x}dx ]
    = 1/3 e^3x sin 2x - 2/3 [1/3 e^3x cos 2x + 2/3 int(sin 2x e^3x) dx]
    = 1/3 e^3x sin 2x - 2/9 e^3x cos 2x - 4/9 I

    Therefore,
    13/9 I = 1/3 e^3x sin 2x - 2/9 e^3x cos 2x

    I= 9/13 [1/3 e^3x sin 2x - 2/9 e^3x cos 2x] + k


    ..typing that was tough, so I'm sorry if I made any typing errors, but this is the sum.


    EDIT: I just typed that whole thing out on my cell phone - how can someone neg me?!
    Probably because you just gave out a full solution, which is frowned upon.
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    (Original post by EEngWillow)
    Probably because you just gave out a full solution, which is frowned upon.
    Wow, we really learn new things every day.

    1) Don't try to help someone. I didn't know I wasn't supposed to give the 'full solution' - I thought that if someone is not quite understanding vague explanations, a solution would help that person to solve similar sums in the future.

    2) People around here just think it's fun to give someone -ve rating. At least I did something to help. Most people just give half-hearted answers and don't give crap if OP actually understood the Hebrew.
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    (Original post by aankhi)
    Wow, we really learn new things every day.

    1) Don't try to help someone. I didn't know I wasn't supposed to give the 'full solution' - I thought that if someone is not quite understanding vague explanations, a solution would help that person to solve similar sums in the future.

    2) People around here just think it's fun to give someone -ve rating. At least I did something to help. Most people just give half-hearted answers and don't give crap if OP actually understood the Hebrew.
    I'm afraid that's not how we do things on here. Maybe take a read of this first.

    It's helpful but for a problem like this one, giving the seed for thought rather than the answer is far more useful. It's a strange application of integration by parts and it would be quite a bit more helpful if the OP evaluated it with a push.

    Secondly, I think you're being disrespectful to those who do give good help on here.
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    (Original post by aankhi)
    Wow, we really learn new things every day.

    1) Don't try to help someone. I didn't know I wasn't supposed to give the 'full solution' - I thought that if someone is not quite understanding vague explanations, a solution would help that person to solve similar sums in the future.

    2) People around here just think it's fun to give someone -ve rating. At least I did something to help. Most people just give half-hearted answers and don't give crap if OP actually understood the Hebrew.
    Someone is in a mood :rofl:
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    (Original post by Farhan.Hanif93)
    I'm afraid that's not how we do things on here. Maybe take a read of this first.

    It's helpful but for a problem like this one, giving the seed for thought rather than the answer is far more useful. It's a strange application of integration by parts and it would be quite a bit more helpful if the OP evaluated it with a push.

    Secondly, I think you're being disrespectful to those who do give good help on here.
    If I could just chip in here (as the OP!). Farhan is right, it is generally better to give hints and guidance rather than solutions straight off. People learn more that way. Though I definitely don't think it's appropriate to neg someone for doing so (especially when they went to the trouble of typing it all out on a phone! (thanks aankhi, for your efforts!)

    Basically, people are incredibly helpful and generous on this (and the other help fora- I try to do my bit to pay it back in Chemistry Help), so let's not be negging each other eh?

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    (Original post by Farhan.Hanif93)
    I'm afraid that's not how we do things on here. Maybe take a read of this first.

    It's helpful but for a problem like this one, giving the seed for thought rather than the answer is far more useful. It's a strange application of integration by parts and it would be quite a bit more helpful if the OP evaluated it with a push.

    Secondly, I think you're being disrespectful to those who do give good help on here.
    I hadn't read that before, so I didn't realize that I shouldn't have posted the solution. At the same time, if words weren't able to explain, only the solution can. I posted the solution because of the post that I quoted before it.

    Secondly, I did not mean to sound disrespectful to people who give good help. It's just that I tried to help and put in effort into it, so the negative rating wasn't very nice.
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    (Original post by aankhi)
    I hadn't read that before, so I didn't realize that I shouldn't have posted the solution. At the same time, if words weren't able to explain, only the solution can. I posted the solution because of the post that I quoted before it.

    Secondly, I did not mean to sound disrespectful to people who give good help. I didn't say that everyone posts randomly. It was said only because it would have been more polite to just inform me nicely that posting solutions isn't the right way to go instead if giving negative rating.
    This forum is not as helpful as it used to be. A few year's ago there was a race to provide help and then someone decided that being helpful was not the 'thing'. The number of people using this forum has consequently declined. I think it is called negative growth.
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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
    Perhaps instead of continuing this discussion about Study Help forum etiquette, the good 'ld days etc, we should ask the OP the slightly more helpful question:

    Do you understand how to evaluate integrals that seem to lead you round in circles now?

    If yes, the method used to get there doesn't particularly matter. The "Don't give full solutions", while a good general guideline is circumventable (imo) in the case where the OP is the kind of person that Plato has always seemed to me - that is, the sort who actually want to understand and not just get the answer.

    On the other hand, If no, then we should probably be endeavouring to explain it further.
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    (Original post by steve2005)
    This forum is not as helpful as it used to be. A few year's ago there was a race to provide help and then someone decided that being helpful was not the 'thing'. The number of people using this forum has consequently declined. I think it is called negative growth.
    What do you feel has gone wrong?

    We can move this discussion to PM if you like because I think this thread has gone off-topic enough as it is.
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    (Original post by EEngWillow)
    Perhaps instead of continuing this discussion about Study Help forum etiquette, the good 'ld days etc, we should ask the OP the slightly more helpful question:

    Do you understand how to evaluate integrals that seem to lead you round in circles now?

    If yes, the method used to get there doesn't particularly matter. The "Don't give full solutions", while a good general guideline is circumventable (imo) in the case where the OP is the kind of person that Plato has always seemed to me - that is, the sort who actually want to understand and not just get the answer.

    On the other hand, If no, then we should probably be endeavouring to explain it further.
    yes, thanks. I have it now. I have two methods, in fact, for this integral. Either the IBP method (now that I recognise that the second step creates something involving the original integral, meaning we can just rearrange and solve - nice!). And secondly Farhan's rather elegant use of de Moivre's formula - (extra nice!).

    God, I heart maths!!

    Thanks for asking, by the way. Appreciate it.

    And thanks for this comment the OP is the kind of person that Plato has always seemed to me - that is, the sort who actually want to understand and not just get the answer. - that is very true of me. I am past the "just passing exams" stage these days. I genuinely want to understand, and the people on this forum are so great at explaining stuff :yep:
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    (Original post by aankhi)
    People around here just think it's fun to give someone -ve rating. At least I did something to help. Most people just give half-hearted answers and don't give crap if OP actually understood the Hebrew.
    Couldn't possibly agree more! I once gave a partial solution to an integral problem after many clueless replies before me, and the post ended getting censored by a section mod (I deleted it then). From that time, I avoid this section (except for today, of course). The rest of TSR is run rather well, but this section is run horribly for some odd reason, which gave it probably the worst quality of any maths forum across the net.
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    (Original post by NCN)
    Couldn't possibly agree more! I once gave a partial solution to an integral problem after many clueless replies before me, and the post ended getting censored by a section mod (I deleted it then). From that time, I avoid this section (except for today, of course). The rest of TSR is run rather well, but this section is run horribly for some odd reason, which gave it probably the worst quality of any maths forum across the net.
    Feel free to suggest ways to improve it rather than just rant about it.
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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 = \Im \left(\displaystyle\int e^{(3+2i)x} dx \right),

    and find that imaginary part.
    He wouldn't get the marks. That is further maths, he can only use IBP in Core3 and 4.
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    (Original post by Aristotle's' Disciple)
    x
    My answer was perfectly valid, thank you - and would have taken as many iterations as your solution.
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    (Original post by Aristotle's' Disciple)
    He wouldn't get the marks. That is further maths, he can only use IBP in Core3 and 4.
    Not true. You can use whatever method you want unless specified.
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    (Original post by Aristotle's' Disciple)
    He wouldn't get the marks. That is further maths, he can only use IBP in Core3 and 4.
    What makes you think he is doing C3 and C4? If you even took a moment to open his sig, you'll see that he's doing an OU course in maths AND if you read the thread, you'd see that he has come across complex numbers/De Moivre's theorem before so it's safe to assume that he'd be allowed to use it. He would only not get the marks if the question states specific use of IBP.
 
 
 
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