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    (Original post by mockel)
    Definitely gave the hint in the paper.
    thats reassuring:rolleyes: - damn edexel - should give hints in the book also - now how about the other quetsions- anyone making any progress on em?

    pk
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    (Original post by Phil23)
    it was integral 1/(x+sqrt(1-x2)) using x=sin t, lol

    dont know how its supposed to help; i got integral of cos t/(sint + cos t), and didn't know where to progress from there:confused:

    pk
    Well, if you divide top and bottom by cost, you get:

    INT 1/(1+tant) dt

    and then try u = tant, you get to:

    INT 1/[(1+u)(1+u2)] du , and then partials...?
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    finally integrate sinx.(1+cos x) / (1-cosx) dx between pi/3 and pi/2
    I = ∫ sinx/(1-cosx) dx + ∫ sinxcosx/(1-cosx) dx

    ∫ sinx/(1-cosx) dx = ln|1-cosx|

    ∫ sinxcosx/(1-cosx) dx = ∫ (1-u)/u du = ln|u| - u = cosx + ln|1-cosx| - 1

    So..
    I = cosx + 2ln|1-cosx| - 1
    Now plug in the limits.

    dont know how its supposed to help; i got integral of cos t/(sint + cos t)
    I'll pass! :P
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    figured out you used sum-product, but HOW did you spot that
    I looked at the answer!

    I(n) = [2sin(n-1)x]/(n-1) + I(n-2)

    The sin[(n-1)x]/(n-1) bit means that cos[(n-1)x] was integrated before.
    The I(n-2) means that we're going to get a sin(n-1)x.

    So I(n)'s nominator should look something like 2cos[(n-1)x]sinx + sin[(n-2)x], so it's just a matter of finding the right identity now.
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    (Original post by mockel)
    Well, if you divide top and bottom by cost, you get:

    INT 1/(1+tant) dt

    and then try u = tant, you get to:

    INT 1/[(1+u)(1+u2)] du , and then partials...?
    prioving very messy, lol - especially when you have to resub back the x in t - are you guys just changing teh limmits to t? - maybe i'll try that and i've getting tan (pi /2) in the final answer!!!! - that proposterous :rolleyes: - where the uni people - mathematical crisis:eek::eek::eek:

    pk
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    (Original post by dvs)
    I = ∫ sinx/(1-cosx) dx + ∫ sinxcosx/(1-cosx) dx

    ∫ sinx/(1-cosx) dx = ln|1-cosx|

    ∫ sinxcosx/(1-cosx) dx = ∫ (1-u)/u du = ln|u| - u = cosx + ln|1-cosx| - 1
    i was getting bogged down using the x=tan (t/2) thing Thanks!
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    (Original post by mockel)
    Well, if you divide top and bottom by cost, you get:

    INT 1/(1+tant) dt

    and then try u = tant, you get to:

    INT 1/[(1+u)(1+u2)] du , and then partials...?
    Ahh.. Now I remember this. It's 3C Q77, right?

    Here you go:
    http://www.thestudentroom.co.uk/t93229.html
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    (Original post by dvs)
    Ahh.. Now I remember this. It's 3C Q77, right?

    Here you go:
    http://www.thestudentroom.co.uk/t93229.html
    thats awesome - cheers!:cool:
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    Oooh, pretty nasty that one. Cool. =)
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    realistically, does anyone reckon that they give those sort of questions in a P5 paper, ever? i've never come across anything like that - then again - i've only done three papers:rolleyes: Also, if they do give stuff like that, they would give sort ort of hint at least wouldn't they, or is there this expectation of further maths students to be creative?

    pk
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    (Original post by Phil23)
    realistically, does anyone reckon that they give those sort of questions in a P5 paper, ever? i've never come across anything like that - then again - i've only done three papers:rolleyes: Also, if they do give stuff like that, they would give sort ort of hint at least wouldn't they, or is there this expectation of further maths students to be creative?

    pk
    I doubt there will be anything like that. Further Maths is only supposed to be of A-Level difficulty really.
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    having probs on this too - aghhh:

    In=∫ (4-x2)n dx, between 0 and 2; have to show that:

    In=8n/(2n+1).In-1

    tried doing (4-x2)(n-1).(4-x2), but not getting anywhere with that, lol - any ideas?
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    Here you go.
    Attached Images
     
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    thanks - but how did you spot to split up the (4-x^2-4) bit - i can't do stull like that, lol,

    cheers

    PK
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    (Original post by Phil23)
    thanks - but how did you spot to split up the (4-x^2-4) bit - i can't do stull like that, lol,

    cheers

    PK
    I thought of it because x^2I(n-1) is obviously useless and the x^2 needs to go somewhere - ie it needs to form one of the terms.
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    Can anyone help me on this?
    Prove that, on the ellipse 9x^2 + 25y^2 = 225 , the normal at P bisects the angle APB, where A & B are the foci and P is any point on the ellipse. (It's Q56 rev ex in P5 heinemann book)
    Kinda nervous about the exam tomorrow!
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    Can anyone help me on this?
    Prove that, on the ellipse 9x^2 + 25y^2 = 225 , the normal at P bisects the angle APB, where A & B are the foci and P is any point on the ellipse. (It's Q56 rev ex in P5 heinemann book)
    Kinda nervous about the exam tomorrow!
    The foci have coords A(5e, 0) and B(-5e, 0). Now find the gradients of the normal, BP and AP:
    m(Normal) = (25y/9x)
    m(BP) = y/(x+5e)
    m(AP) = y/(x-5e)

    The formula for the angle t between two lines with gradients p and q is:
    tan(t) = (p-q)/(1+pq)

    We want the angles between PB and the normal and PA and the normal to be equal, i.e. we want:
    [m(pb) - m(n)]/[1 + m(n)m(pb)] = [m(pa) - m(n)]/[1 + m(n)m(pa)]
    ([m(pb) - m(n)][1 + m(n)m(pa)])/([1 + m(n)m(pb)][m(pa) - m(n)]) = 1

    Enjoy simplifying!

    There's also a MUCH better method if you're familiar with P6 vectors, mainly the part about finding an angle between two lines.
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    Hey, I need some help on this reduction question...

    In = ∫ xn.√(1-x) dx , from 0 to 1

    The first part was show that:
    In = [2n/(2n+3)]In-1 , n>1

    That part's fine.

    The next part was:

    Hence, find ∫ (cos5θ)(sin²θ) dθ , from 0 to pi/2

    I tried various substitutions. For example, sinθ = x, cosθ = x, sin²θ = x, sin4θ = x...because I was trying to get the limits of 0 to 1. Fair enough, the limits are fine, but I couldn't get into a position to be able to use the reduction formula proved in the first part.

    Any help appreciated :)
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    (Original post by mockel)
    In = ∫ xn.√(1-x²) dx , from 0 to 1

    The first part was show that:
    In = [2n/(2n+3)]In-1


    Hence, find ∫ (cos5θ)(sin²θ) dθ , from 0 to pi/2
    x=cos\theta\\

dx=-sin\theta d\theta\\

d\theta=-cosec\theta dx\\

\theta = 0,x=1\\

\theta = \frac{\pi}{2},x=0\\
    We have \int_1^0 -x^{5} sin\theta dx\\

=\int_1^0 -x^{5} (1-x^{2})^{\frac{1}{2}} dx\\

=I_{5}
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    god how did you do the first part??
 
 
 
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