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    (Original post by DJMayes)
    Agreed. I prefer lower case omega (the curly w to which you refer) more than upper case as well.

    What modules are you actually sitting this summer?
    It's so cool that you're starting Cambridge this September. Will you appear on TSR after that new chapter in your life begins DJ?
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    (Original post by reubenkinara)
    It's so cool that you're starting Cambridge this September. Will you appear on TSR after that new chapter in your life begins DJ?
    No he will vanish into thin air :lol:
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    (Original post by DJMayes)
    Agreed. I prefer lower case omega (the curly w to which you refer) more than upper case as well.

    What modules are you actually sitting this summer?
    C3/4, M2, Electrons, Waves and Photons, some chem exam, and STEP I
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    (Original post by Robbie242)
    No he will vanish into thin air :lol:
    I'm actually really surprised people like nuodai and Farhan.Hanif had time to moderate TSR whilst doing a Cambridge Maths degree.
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    (Original post by Robbie242)
    No, he will vanish into thin air :lol:
    I wouldn't be surprised. He is a demigod after all!
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    (Original post by Felix Felicis)
    I'm actually really surprised people like nuodai and Farhan.Hanif had time to moderate TSR whilst doing a Cambridge Maths degree.
    Maybe it isn't all that demanding after all.

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    (Original post by Felix Felicis)
    I ****ing execrate all of you

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    I = \displaystyle\int \dfrac{dx}{(x+q) \sqrt{ax^{2} + bx + c}}

    Let x + q = t^{-1} \Rightarrow I = - \displaystyle\int \dfrac{dt}{\sqrt{t^{2}} \sqrt{ax^{2} + bx + c}}

    = - \displaystyle\int \dfrac{dt}{\sqrt{(aq^{2} - bq + c) t^{2} + (b-2aq)t + a}}

    Let aq^{2} - bq + c = \alpha, b - 2aq = \beta (yes it's cheating, so bite me :teehee: )

    I = - \displaystyle\int \dfrac{dt}{\sqrt{ \alpha^{2} t^{2} + \beta t + a}}

    = - \dfrac{1}{\sqrt{ \alpha }}\displaystyle\int \dfrac{dt}{\sqrt{ \left[t + \frac{\beta}{2 \alpha} \right]^{2} + \frac{a}{\alpha} - \frac{\beta^{2}}{4 \alpha^{2}}}}

    Let  \frac{a}{\alpha} - \frac{\beta^{2}}{4 \alpha^{2}} = \gamma (bite me :teehee:)

    Let \left[t + \frac{\beta}{2 \alpha} \right] = \gamma \tan u

    \Rightarrow I = - \dfrac{1}{ \sqrt{\alpha}} \displaystyle\int \sec u du

    = - \dfrac{1}{ \sqrt{\alpha}} \ln (\sec u + \tan u ) + \mathcal{C}

     = - \dfrac{1}{ \sqrt{\alpha}} \ln \left( \sec \left( \dfrac{ [t + \frac{ \beta}{2 \alpha} ]}{\gamma} \right) + \left( \dfrac{ [t + \frac{ \beta}{2 \alpha} ]}{\gamma} \right) \right) + \mathcal{C}

     = - \dfrac{1}{\sqrt{\alpha}} \ln \left( \left( \dfrac{ [t + \frac{\beta}{2 \alpha} ]}{\gamma} \right) + \sqrt{1 + \dfrac{ [ t + \frac{\beta}{2 \alpha} ]^{2}}{\gamma^{2}}} \right)  + \mathcal{C}

    where t = x + q, \alpha = aq^{2} - bq + c, \beta = b - 2aq, \gamma = \dfrac{a}{\alpha} - \dfrac{\beta^{2}}{4 \alpha^{2}}

    Who amongst you are courageous enough to endeavour in the righteous quest of checking my working? :teehee:
    Whether it's a tex issue or not it appears an \alpha^2 has crept in just after you stated what \alpha was :lol:
    The \frac{1}{t} was a clever step
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    (Original post by joostan)
    Whether it's a tex issue or not it appears an \alpha^2 has crept in just after you stated what \alpha was :lol:
    The \frac{1}{t} was a clever step
    Damn yeah, tex :teehee: I'm surprised you bothered to check it :rofl: Yeah, the t^{-1} substitution is what cracks integrals like this

    I like to call this type of proof...proof by intimidation
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    (Original post by Felix Felicis)
    Damn yeah, tex :teehee: I'm surprised you bothered to check it :rofl: Yeah, the t^{-1} substitution is what cracks integrals like this

    I like to call this type of proof...proof by intimidation
    I gave it a brief glance - tbh it just jumped out at me.
    I've not seen one like that
    I did a STEP q with \displaystyle\int \dfrac{dx}{\sqrt{x^2+2bx+c}}
    and another with a \dfrac{1}{u} substitution, but that is cheeky. :giggle:
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    (Original post by joostan)
    I gave it a brief glance - tbh it just jumped out at me.
    I've not seen one like that
    I did a STEP q with \displaystyle\int \dfrac{dx}{\sqrt{x^2+2bx+c}}
    and another with a \dfrac{1}{u} substitution, but that is cheeky. :giggle:
    I think I remember that! Not sure which paper though, but it's a useful sub :teehee: I wonder if there's some place which just has a list of useful integral substitutions :lol:
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    (Original post by Felix Felicis)
    I wonder if there's some place which just has a list of useful integral substitutions :lol:
    I would love to see that list
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    61-C2
    63-C1
    58-S1
    Any chance of an A?
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    (Original post by Felix Felicis)
    I think I remember that! Not sure which paper though, but it's a useful sub :teehee: I wonder if there's some place which just has a list of useful integral substitutions :lol:
    Make one.
    A brief internet search shows up a few - mainly trig ones though :cool:
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    (Original post by Shady778)
    61-C2
    63-C1
    58-S1
    Any chance of an A?
    I'd have thought that would be a low A, rather than a high B - depends on the year and exam board of course
    EDIT: As MathsNerd says that is raw score right?
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    (Original post by Shady778)
    61-C2
    63-C1
    58-S1
    Any chance of an A?
    Is that marks or UMS? I'm not quite sure that could be high enough but you'll just have to wait and see what the grade boundaries come out as


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    (Original post by tigerz)
    lmao seems like something you'd do, this means I can casually make:
    \sqrt{a^2+b-\frac{c^3}{7}}= or Random Maths Shiz=
    I will do so if required when joostan gives me that problem
    For hidden quadratics a friend of mine insisted on using the union jack :lol:
    Haha - as if I'd give you such an easy problem :laugh:
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    I'm kidding
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    (Original post by MathsNerd1)
    Is that marks or UMS? I'm not quite sure that could be high enough but you'll just have to wait and see what the grade boundaries come out as


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    This is raw marks nor percentage or ums
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    (Original post by reubenkinara)
    It's so cool that you're hopefully starting Cambridge this September. Will you appear on TSR after that new chapter in your life begins DJ?
    Fixed the quote. I don't see any reason why not, although it may well be less than currently. Either way, I have to get there first!
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    (Original post by Shady778)
    This is raw marks nor percentage or ums
    Just about an A, but judging on these papers for this exam session (if this is edexcel) I'm afraid to say that seems like a very high B grade

    You need 240/300 ums for an A overall at AS level maths

    I predict that C1 will be 63/75 for an A
    I predict that C2 will be 62/75 for an A
    I predict that S1 will be 61/75 for an A

    Each unit is given a ums mark out of 100
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    (Original post by Robbie242)
    Just about an A, but judging on these papers for this exam session (if this is edexcel) I'm afraid to say that seems like a very high B grade
    thanks anyway
 
 
 
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