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    Could anyone please help me with these three questions please, I did try getting the Mock Mark Schemes but no-one had them, so any help whatsoever would be mucho appreciated .

    On this first one, I'm not sure what substitution to use, I've tried using integration by parts from the start but I get a really messed up answer.

    Question 1 out of 3 is done thanks to all of you, please could you help with the other two also and now question 2 also thanks solely to Jonny. Just one question to go, I hope you can help, but even if you all can't, then thanks anyway you've helped me so much already.
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    On this one I've tired it so many times, but always end up with the wrong answer, I always get (41/25) in the ^(3/2) bracket and sometimes its the ^(-3/2) bracket
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    I had a go, but I'm afraid my P1 knowledge doesn't cover that!
    Looks tricky, I'd try factorising something, can you substitute a symbol for dx? Then split the fraction up and simplify, taking terms upto the top and then intergrating?

    I really don't know I'm sorry, that looks pretty hard!
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    And finally this one, I just have no idea how to convert from intrinsic to cartesian at all :confused: .
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    (Original post by mik1a)
    I had a go, but I'm afraid my P1 knowledge doesn't cover that!
    Looks tricky, I'd try factorising something, can you substitute a symbol for dx? Then split the fraction up and simplify, taking terms upto the top and then intergrating?

    I really don't know I'm sorry, that looks pretty hard!

    You made me laugh and gave me an idea, thanks. I'll try using x+3 as a substitution and then change the x+2 into u-1 and then split the fraction up. Thanks .
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    Lol... I have no idea what that does but glad to be of service
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    by separating the fraction, and completing the square you get
    x/(root(x+3)^2-(root5)^2) and 2/(root(x+3)^2-(root5)^2)

    the integral of the second bit is 2arcosh [(x+3)/root5]
    the first bit... am still looking at but substitution may work! :confused:
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    (Original post by ogs)
    by separating the fraction, and completing the square you get
    x/(root(x+3)^2-(root5)^2) and 2/(root(x+3)^2-(root5)^2)

    the integral of the second bit is 2arcosh [(x+3)/root5]
    the first bit... am still looking at but substitution may work! :confused:

    Thanks , thats how I originally tried it, our teacher taught us to always complete the square on the root first, but I wasn't sure about the first fraction either
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    (Original post by mik1a)
    Lol... I have no idea what that does but glad to be of service
    Glad to be in reciept of servive .
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    (int) (x + 2) / sqrt[x^2 + 6x + 4] dx
    = (int) (x + 2) / sqrt[(x + 3)^2 - 5] dx
    = (int) (u - 1) / sqrt[u^2 - 5] du . . . . . (substituting u = x + 3)
    = (int) u / sqrt[u^2 - 5] du - (int) 1 / sqrt[u^2 - 5] du
    = sqrt[u^2 - 5] - arccosh[u / sqrt(5)] + constant
    = sqrt[(x + 3)^2 - 5] - arccosh[(x + 3) / sqrt(5)] + constant
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    (Original post by Jonny W)
    (int) (x + 2) / sqrt[x^2 + 6x + 4] dx
    = (int) (x + 2) / sqrt[(x + 3)^2 - 5] dx
    = (int) (u - 1) / sqrt[u^2 - 5] du . . . . . (substituting u = x + 3)
    = (int) u / sqrt[u^2 - 5] du - (int) 1 / sqrt[u^2 - 5] du
    = sqrt[u^2 - 5] - arccosh[u / sqrt(5)] + constant
    = sqrt[(x + 3)^2 - 5] - arccosh[(x + 3) / sqrt(5)] + constant
    Thanks Jonny, mik1a gave me the idea to do that, and I just got that answer as well, so hopefully that means we're both correct, cool I can cross one of the questions off the list now .
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    The ROC question

    The trick is to substitute x = k/2 as soon as possible.

    Let f(x) = k*arctan(x/k).

    Then f'(x) = 1 / (1 + x^2/k^2). So f'(k/2) = 4/5.

    Also f''(x) = -2x / [k^2 (1 + x^2/k^2)^2]. So f''(k/2) = -16/(25k).

    So

    ROC at x = k/2
    = (1 + f'(k/2)^2)^(3/2) / |f''(k/2)|
    = (1 + 16/25)^(3/2) * 25k/16
    = (41/25) (41/25)^(1/2) (25/16) k
    = (41/16) (41/25)^(1/2) k
    = (41/80) 41^(1/2) k
    = 41^(3/2) k / 80.
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    (Original post by Jonny W)
    The ROC question

    The trick is to substitute x = k/2 as soon as possible.

    Let f(x) = k*arctan(x/k).

    Then f'(x) = 1 / (1 + x^2/k^2). So f'(k/2) = 4/5.

    Also f''(x) = -2x / [k^2 (1 + x^2/k^2)^2]. So f''(k/2) = -16/(25k).

    So

    ROC at x = k/2
    = (1 + f'(k/2)^2)^(3/2) / |f''(k/2)|
    = (1 + 16/25)^(3/2) * 25k/16
    = (41/25) (41/25)^(1/2) (25/16) k
    = (41/16) (41/25)^(1/2) k
    = (41/80) 41^(1/2) k
    = 41^(3/2) k / 80.
    You're a godsend Jonny, I got the two derivatives correct, but I screwed up the second half of the problem everytime I tried it, thanks so much for your help, another question done.

    I've read something about reputation on here and if I work out how, I'll try and raise yours or whatever .
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    I'm working on the intrinsic coordinate one at the moment. They are nasty things.
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    (Original post by chrisbphd)
    I'm working on the intrinsic coordinate one at the moment. They are nasty things.
    Thanks Chris . They are really nasty, we haven't even been taught them properly, otherwise I'd have at least known the method to convert back to cartesian and our teach reckons they're much better than cartesian coordinates too :rolleyes: .
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    I've done the first part of this last question to but can't seem to do the locus part, so if anyone could again help, sorry for all these questions .

    For the first part I got the answer:

    y = (p^2)x - 2p^3 +2p^(-1)
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    (Original post by Morgan)
    I've done the first part of this last question to but can't seem to do the locus part, so if anyone could again help, sorry for all these questions .

    For the first part I got the answer:

    y = (p^2)x - 2p^3 +2p^(-1)
    You are correct for the first part. I'm doing the locus now. No luck on the intrinsics, they don't appear on any AQA B pure modules (1-7) and I did them all. I tried though and got about half way through before everything collapsed.

    I'll have another go if I have time later.
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    The method for the second part is as follows:

    1. Determine the position of point P.
    2. Using elementary co-ordinate geometry, find the mid point of PQ.
    3. Work out the locus of M as P varies.

    Given that there are 8 marks for this question, i expect that they would me in the following ratios:

    1. 2 marks
    2. 1/2 marks
    3. 4/5 marks
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    Thanks Chris, but I still don't get how to do the locus question, any chance you could enlighten me. Thanks eitherway though .
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    (Original post by Morgan)
    Thanks Chris, but I still don't get how to do the locus question, any chance you could enlighten me. Thanks eitherway though .
    basically the locus of a point is how does that point vary as the parameter varies,so lets take a hypothetical point P which has coordinates (bsinx,acosx) therefore x=bsinx and y=acosx.Now what u have to do for the locus of P is eliminate the parameter.
    (x/b)^2=sin^2(x) and (y/a)^2=cos^2(x)
    cos^2(x)+sin^2(x)=(y/a)^2+(x/b)^2=1
    so the locus of P is (y/a)^2+(x/b)^2=1, understand?
 
 
 
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