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# P5 questions help, thread:) watch

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

pk
2. (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

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...?
3. 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
4. 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.
5. (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 - where the uni people - mathematical crisis

pk
6. (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!
7. (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
8. (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!
9. Oooh, pretty nasty that one. Cool. =)
10. 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 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
11. (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 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.
12. 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?
13. Here you go.
Attached Images

14. 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
15. (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.
16. 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!
17. 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.
18. 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 :)
19. (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

We have
20. god how did you do the first part??

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