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    anyone got this paper; i'm still frustrated i messed it up and want to do it for fun so i can like make sure i can do the questions now

    if you have it, send it to me

    cheers!

    phil
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    http://www.thestudentroom.co.uk/show...&postcount=122
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    thanks for that...i didn't do as bad as i first thought...thought i got 65% initially, but looking back at it, i've figured i got ~82%

    does anyone have the copy of the P6 paper for this summer?

    cheers again

    PK
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    And sorry for hijacking, does anyone have the P4 paper (June 05)? I want to add up the marks I'm likely to get... Cheers!
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    (Original post by RobbieC)
    And sorry for hijacking, does anyone have the P4 paper (June 05)? I want to add up the marks I'm likely to get... Cheers!
    oh that paper was piss - sure you'll get 100%
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    anyone know how to integrate x^2 / sqrt(1-x^2))?
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    (Original post by Phil23)
    anyone know how to integrate x^2 / sqrt(1-x^2))?
    Have you tried x=sinu?
    It looks like that transforms it into \int \sin ^{2} u du which is easy to integrate.
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    (Original post by Gaz031)
    Have you tried x=sinu?
    It looks like that transforms it into \int \sin ^{2} u du which is easy to integrate.
    i used a hyperbolic substitution in the exam, lol - no wonder i was going round in big complicated circles

    tried the sinuthing but not getting it cos when you change the limits, 2=sinu is undefined...its from that P5 paper - anyone know how do do that integral...of xarcoshx between 1 and 2?
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    xarcoshx between 1 and 2?
    x=\cosh u \\

dx=\sinh u du \\

\int_{1}^{2} xarcoshx dx = \int_{0}^{arcosh2} u\cosh u \sinh u du \\

=\frac{1}{2}\int_{0}^{arcosh2} usinh2u du \\

v=u,\frac{dv}{du}=1,\frac{dt}{du  }=sinh2u,t=\frac{1}{2}cosh2u \\

\int_{1}^{2} xarcoshx dx =\frac{1}{2}[ [\frac{1}{2}ucosh2u]_{0}^{arcosh2} - \int_{0}^{arcosh2} \frac{1}{2} cosh2u du ] \\

=\frac{1}{4}(arcosh2)cosh(2arcos  h2) - [\frac{1}{8}sinh2u]_{0}^{arcosh2}

=\frac{7}{4}ln(2+\sqrt{3})-[\frac{\sqrt{3}}{2}] \\

=\frac{7}{4}ln(2+\sqrt{3})-\frac{\sqrt{3}}{2} \\

    The above was one way to do it. However, like you I started integration by parts before a substitution so my method in the exam was slightly messier, though it gave the answer.

    The substitution for the term you had from integration by parts wouldn't have worked well due to the limits. Sorry about that. I wonder whether it would have actually worked though if we had found the value of arcsin2 using complex numbers and then carried out the integration and evaluation using limits as normal.

    How come you're doing this paper again? Why not learn something new if you're in the mood to do something productive?
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    i didn't get that qeustion in the exam, and still not getting it - and i wont be able to rest until i get it, lol...

    i did most of what you've posted but i couldn't simplfy that bit with the cosh thing in the sinch expression...think i got 5/10 marks for it...gd enough - more than made up for it in my other modules

    phil...might start working from that pack oxford sent me...got lots of problem sheets in the summer
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    (Original post by Phil23)
    i didn't get that qeustion in the exam, and still not getting it - and i wont be able to rest until i get it, lol...

    i did most of what you've posted but i couldn't simplfy that bit with the cosh thing in the sinch expression...
    Just use your calculator and square the answer. If the square is a rational number then you just know you've got a root.
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    This thread took a delightfully complicated turn... And the P4 paper... 100%?? Heh, funny.

    More like 40% if im lucky. Im hoping for a 50% secretely though!!!
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    (Original post by Nima)
    x = sinu. this is simple to spot, you'll end up getting cosu as the denominator and things progress from there.
    read the other posts - the limits to the quetsion aer a bit awkward and would involve complex roots, so there's got to be another substitution - i used a hyperbolic one, but it got really messy:confused:
 
 
 
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