You are Here: Home

# P3 Revision topic - 1:15pm 9th June - OCR/Edexcel watch

1. ∫ ² ³ ± ÷ ° √ ∂ ∞ ≈ ≠ ≡ ≥ ≤ ø Ø ½ ¼ « » µ π

*******s, why oh why did I volunteer myself to write notes on P3, I must be a saint or something, anyway if you care to show your appreciation by providing me with gems/rep, then I won't complain.

Ok, first off, I've been going through a shedload of papers and have gotten used to the sort of questions they ask, below are just some notes to supplement the notes you already/should have, it's not to replace your existing methods/practices so don't blame me if you don't know what I'm blabbing about, if you get stuck or don't understand something then ask.

Trigonometry and Identities

You are expected to know:

tan^[email protected]+1 = sec^[email protected]
sin^[email protected]+cos^[email protected]=1
[email protected]/[email protected][email protected]

A big one:

[email protected]=cos^[email protected]^[email protected] for this one is important if you need to integrate either cos^[email protected] or sin^[email protected]

The '[email protected]=cos^[email protected]^[email protected]' can be rewritten
by simply substituting either 'cos^[email protected]' with '1-sin^[email protected]' or substituting 'sin^[email protected]' with '1-cos^[email protected]', thus producing:

[email protected]=cos^[email protected]^[email protected] = [email protected]=cos^[email protected](1-cos^[email protected]) OR
[email protected]=cos^[email protected]^[email protected] = [email protected]=(1-sin^[email protected])-sin^[email protected]

which rearranged can yield:

cos^[email protected]=([email protected]+1)/2 OR
sin^[email protected]=([email protected])/2

What this means is that it will allow you to integrate sin^[email protected] or cos^[email protected]

You also need to know the expansions of:

cos2x = cos(x+x)
sin2x = sin(x+x)
tan2x = tan(x+x)

the expansions are in the forumula book and look something like:

cos(A+B) = cosAcosB-SinASinB
cos(A-B) = cosAcosB+SinASinB
Sin(A+B) = SinACosB+CosASinB
Sin(A-B) = SinACosB-CosACosB
Tan(A+B) = (TanA+TanB)/(1-TanATanB)
Tan(A-B) = (TanA-TanB)/(1+TanATanB)

(notice how cos and tan have opposite signs)

You just replace the 'A' and 'B' with 'X' fairly straight forward, all you need is common sense.

Integration/Differentiation of Trig

You should know:

d/dx of sinx = cosx
d/dx of cosx = -sinx
d/dx of tanx = sec^2x

also:

d/dx of sinmx where 'm' is a constant is mcosmx
d/dx of cosmx where 'm' is a constant is -msinmx

e.g d/dx of sin4x = 4cos4x

∫sinx dx = -cosx
∫cosx dx = sinx
∫sec^2x dx = tanx

also:

∫sinmx where 'm' is a constant is (-cosmx)/m
∫cosmx where 'm' is a constant is (sinmx)/m

Well how exactly do we integrate sin^2x or cos^2x? Well remember what you read before about the identities?

sin^[email protected]=([email protected])/2
cos^[email protected]=([email protected]+1)/2

Well you just integrate them, for example to integrate

sin^[email protected], we know that sin^[email protected]=([email protected])/2 therefore we just integrate ([email protected])/2

∫sin^[email protected] [email protected] is the same as 1/2∫([email protected]) which is:

[@/2 - ([email protected])/4 + C]
2. Binomial Expansion

Nice easy topic. This is used to expand brackets where the power is non-integer or negative non integers:

e.g

(1+5x)^-1/2

A general binomial is:

(a+x)^n

The formula is 1+nx+[(n(n-1))/2!](x)^2+[(n(n-1)(n-2))/3!](x)^3..........and so on. You get the picture. From my experience of the past papers, the most I have seen a question ask is for is the expansion up until x^3 terms, they are typically up to x^2.

e.g. Expand (1+4x)^-1/2 up until X^2:

General forumula is 1+nx+[(n(n-1))/2!](x)^2...................therefor e sub in the values:

n=-1/2
x=4x

= 1+(-1/2)(4x)+[((-1/2)(-3/2))/2](4x)^2
= 1 - 2x + 6x^2

Sometimes they will ask you to write down the numbers for which the expansion is valid. Its always [x]<1, (where [] is modulus) in this case x=4x therefore its [4x]<1 which is [x]<1/4.

In some cases, they may ask you to expand somethin such as:

(4+8x)^1/2, for us to expand the bracket the number '4' has to be a '1', so what we do is take the number '4' outside of the bracket and raise it to the same power as the bracket:

4^1/2 (4+8x)^1/2

and divide everything in the bracket by '4'

4^1/2 (1+2x)^1/2 = 2(1+2x)^2

You would then expand the bracket as per normal and then multiply everything by two, i.e.

2(1+nx+[(n(n-1))/2!](x)^2...................)

THERE IS NO WAY YOU CAN EXPAND A BINOMIAL IF THE CONSTANT IN THE BRACKET ISN'T ONE! SO YOU HAVE TO DO THIS!!!
3. Differentiation

Differentiation, well what can I say, but differentiation. You neeed to know your trig identities, like I showed you before, you also need to know basic trig, which I won't show you how to do, as it would be insulting. You need to be familiar with the chain rule.

dy/dx = dy/du * du/dx

If you think about it, it's like saying: 1 = 2/3 * 3/2

You are usually given an equation such as:

Y = (4+6x)^5, well how do you differentiate. Well you say that 'u' = (4+6x) and rewrite 'Y' in terms of 'u'. i.e.

Y = (4+6x)^5 becomes Y=u^5. So you are left with two equations:

Y=U^5 and U=(4+6x), differentitae both:

dy/du = 5u^4
du/dx = 6

therefore dy/dx = dy/du * du/dx = 30u^4 and just replace 'u' with (4+5x), giving:

Y = 30(4+6x)^4.

You can extend this to trig. E.g. Diffentiate (Sinx)^5

Y = u^5
dy/du = 5u^4

U=sinx
du/dx=cosx

dy/dx = cosx * 5u^4

= 5cosx(sinx)^4

Product Rule

The product rule is:

d/dx (uv) = u(dv/dx) + v(du/dx).

And is used to differentiate something multiplied by something else. E.g. Differentiate sinxcosx

say that

u=sinx
du/dx=cosx
v=cosx
dv/dx=-sinx

d/dx (sinxcosx) = sinx(-sinx) + cosx(cosx) = cos^2x-sin^2x

Simple. But this isn't in the formula booklet, and you need to learn it off by heart.

Quotiant rule

The quotiant rule is used to differentiate an equation of the form (a/b) where you have a fraction:

It is expressed as:

d/dx (u/v) = [v(du/dx)-u(dv/dx)]/v^2

So if I was asked to differentiate tanx, I would split it into sinx/cosx (you should know your trig identites, see.)

d/dx (sinx/cosx)

u=sinx
du/dx=cosx
v=cosx
dv/dx=-sinx

d/dx (sinx/cosx) = [cosx(cosx)-sinx(-sinx)]/(cosx)^2

=cos^2x+sin^2/cos^2x we know that cos^2x+sin^2x =1 so its 1/cos^2x = sec^2x, or 1+tan^2x which we know is sec^2x.
4. ∫ ² ³ ± ÷ ° √ ∂ ∞ ≈ ≠ ≡ ≥ ≤ ø Ø ½ ¼ « » µ π

Integration

Oh *******y *******s, I gather here that nobody really likes integration, its not that it is hard its just that there are so many ways of going wrong when your integrating.

Right let's begin, you are expected to know the basic principles of integration, all that add one to power and stuff, and no, I shan't elaborate on how to do it, as you should know already, if you really want me to explain it to you then you shouldn't be anywhere near a P3 paper.

There are two ways of integrating, integration by parts and integration by substitution. I shall explain both of them with examples.

Integration by parts

The formula is:

∫ u(dv/dx) = uv-∫ v(du/dx)

You are expected to know this off by heart, its not in the formula booklet.

Its used to integrate an equation where two things are multiples together, and is the same formula used to differentiate but rearranged, it should therefore look familiar.

Example

∫(x+1)e^-x dx

The fomula is

∫u(dv/dx) = uv-∫ v(du/dx) compare with ∫(x+1)e^-x dx

With integration by parts, the 'u' and 'dv/dx' correspond to the (x+1) and e^-x, what this means is that one of the two will be 'u' and the other 'dv/dx', don't ask why, its just a case of where it has to be.

Your job is to select the one which is which. So:

∫u(dv/dx) = uv-∫ v(du/dx) compare with ∫(x+1)e^-x dx

you select 'u' to the one which will differentiate to its most simplest. So lets differentiate.

d/dx of (x+1) = 1
d/dx of (e^-x) = -e^-x

from this we can summise that (x+1) simplifies to the most simplest. Therefore:

u=(x+1) and dv/dx=e^-x, you now have to differentiate 'u' and integrate 'dv/dx'.

u=(x+1)
du/dx=1

dv/dx=e^-x
V=-e^-x

then bung them into the formula:

∫u(dv/dx) = uv-∫ v(du/dx)

(This means, the integral of the left handside is uv-∫ v(du/dx))

∫u(dv/dx) = (x+1)(-e^-x)-∫ (-e^-x)(1)

You then have to integrate the ∫ (-e^-x)(1) which is just e^-x. Therefore:

(x+1)(-e^-x)-∫ (-e^-x)(1) becomes (x+1)(-e^-x)-e^-x + C

Simple, no? However there are special cases in which 'u' isn't the simplest.

Consider:

∫xlnx dx

I have said that 'u' is the one that you chose that differentiates to its most simplest, right?

d/dx of x = 1
d/dx of lnx = 1/x

In this case u=x would be the best choice, right? Wrong! it has to be lnx. I'll explain, so lets assume I chose u=x:

u=x
du/dx=1

dv/dx=lnx
v=Here is the problem, we can't integrate lnx, you can only differentiate lnx.

This is why u=lnx. Well, not one for leaving a job half done, lets finish this equation.

u=lnx
du/dx=1/x

dv/dx=x
v=(x^2)/2

∫u(dv/dx) = uv-∫ v(du/dx)

(x^2lnx)/2 - ∫[(x^2)/2](1/x)

= (x^2lnx)/2 - (x^2)/4 + C

Integration by substitution

I actually like this method. Let's begin. In an exam question you are always told what to use as the substitution, I think the only way I can explain this is by jumping straight into a question:

Using u=lnx as the substitution, ∫(lnx)/x dx.

First thing is to differentiate u, therefore du/dx = 1/x, once you have done that rearrange the du/dx to make dx the subject:

dx= xdu sub this in where dx is therefore:

∫(lnx)/x dx = ∫(lnx)/x xdu the 'x' cancels leaving:

∫(lnx) du. Our original substitution was u=lnx therefore we can simply replace the lnx with 'u'.

∫u = (U^2)/2+c and we simply put the lnx where 'u' us leaving:

[(lnx)^2]/2 + C

∫ ² ³ ± ÷ ° √ ∂ ∞ ≈ ≠ ≡ ≥ ≤ ø Ø ½ ¼ « » µ π

Backed by popular demand I bring you:

Vectors

*drum roll*

What can I say, apart from vectors are gay. They are actually quite easy once you understand them, and I am gonna explain why.

We'll start off easy:

Two points A and B have position vectors 2i-j+2k and 2j+3k respectively:

i)Find the vector AB, and hence find the length of AB

ii)Find a vector equation for the line through A and B.

What I normally do is write out A and B in columm form, i.e:

i=x
j=y
z=k

would correspond to each other, cause that is all they mean:

therefore vector A would be:

(2 ) 2i
(-1) -j
(2 ) 2k

B would be:

(0 ) 0i
(2 ) 2j
(3 ) 3k

Now we have the vectors in nice easy forms we can begin the questions:

i)Find the vector AB, and hence find the length of AB

You have to visualise where A and B are, the co-ordinates are how far away they are from the origin also called 'O' (0i+0j+0k). Therefore to get from the origin to A, its (2i-j+2k), this means '+2' across the X axis, '-1' along the 'Y' axis and '+2' along the 'Z' axis. Still with me? Good. And to get back from 'A' to 'O' its the opposite signs, i.e. -1*(2i-j+2k) = (-2i+j-2k), this means '-2' across the X axis, '+1' along the 'Y' axis and '-2' along the 'Z' axis.
Well that's vectors at a glance.

From what I have explained above. which any simpeton should understand vectors A and B are positioned relative to the origin. So we can call them OA and OB, with OA meaning how to get from O to A.

The question is asking us how to get from A to B basically, and you have to work out how to do so using ONLY the information provided.

To get from A to B, you have to think about how to get from the point 'A' to the point 'B'. From the information, we only know how to get from the Origin to A, and how to get from the Origin to B, right? OA and OB.

To get from A to O, its simply -OA, and to get to be its OB right, therefore:

-OA+OB=AB OR:
OB-OA=AB

Therefore if OA is:

(2 )
(-1)
(2 )

and OB:

(0 )
(2 )
(3 )

and OB-OA=AB then:

(0 ) - (2 ) = (-2)
(2 ) - (-1) = (3 )
(3 ) - (2) = (1 )

The position vector AB can be expressed as:

(-2)
(3 )
(1 )

or (-2j+3k+l)

See wasn't so hard now, was it?

i)Find the vector AB, and hence find the length of AB

For the length, its the square root of the sum of the vector squared, aslo called the modulus (the modulus has a different meaning than what you think) , i.e:

√[(i^2+j^2+k^2)]

in this case it would be:

√[(-2)^2+(3)^2+(1)^2] = √14

√14 is the length of AB.

ii)Find a vector equation for the line through A and B.

Vector equations are always in the form:

r=a+tp

a=(any point that lies on the line)
t=a letter
p=(a line that is parallel to the vector)

Thus in our example, there are two known points on our line, 'A' and 'B'. so we could use any of those for 'a'.

For P, we just use the vector of AB as its parallel to the line AB (actually they are both one and the same). Therefore if we all these values in, we get:

____(2 ) +t(-2)
r = (-1) +t(3 )
____(2 ) +t(1 )

or

r = (2i-j+2i)+t(-2i+3j+i)

I'll give you a rep tomorrow to reward you for your hard work!!! , i can't hive you one at the moment!!!
6. d/dx of tanx = -lncosx
d/dx of tan^2x = sec^2x

isnt d/dx of tanx = sec^2x
d/dx of tan^2x = 2sec^2 2x
7. (Original post by Jubba)
d/dx of tanx = -lncosx
d/dx of tan^2x = sec^2x

isnt d/dx of tanx = sec^2x
d/dx of tan^2x = 2sec^2 2x
Yeah, **** it is, I mean the integral of tanx = -lncosx
d/dx of tanx = sec^2x

What can I say, at least you know that it is me who did this, and I'm not simply copying off a website.
8. (Original post by Bhaal85)
Yeah, **** it is, I mean the integral of tanx = -lncosx
d/dx of tanx = sec^2x

What can I say, at least you know that it is me who did this, and I'm not simply copying off a website.

isnt the integral of tanx = ln sec x?
9. (Original post by Jubba)
isnt the integral of tanx = ln sec x?
Wait:

sinx/cosx dx u=cosx du/dx = -sinx therefore dx = -du/sinx

integral of sinx/cosx -du/sinx = -lnu = -lncosx
10. Good work, must have taken some time, i've added to your rep.
11. (Original post by Bhaal85)
Wait:

sinx/cosx dx u=cosx du/dx = -sinx therefore dx = -du/sinx

integral of sinx/cosx -du/sinx = -lnu = -lncosx
-lncosx = ln(cosx)^-1

=ln(1/cosx)
=lnsecx

So theyre the same basically..
12. (Original post by ShOcKzZ)
Good work, must have taken some time, i've added to your rep.
Cheers dude. All these notes are from memory, might I add.
13. I didnt know it was in the afternoon - gay
14. (Original post by imasillynarb)
I didnt know it was in the afternoon - gay
At least you get more sleep, and can revise a bit more in the morning.
15. (Original post by Bhaal85)
At least you get more sleep, and can revise a bit more in the morning.
Thats the problem, if I revise in the morning I dont actually want to, just forcing myself to, and if I get anything wrong whilst revising Ill not be feeling good
16. (Original post by imasillynarb)
Thats the problem, if I revise in the morning I dont actually want to, just forcing myself to, and if I get anything wrong whilst revising Ill not be feeling good
Oh no, dare I say it, your turning into AT.

Can't you go into college for some support in the morning?
17. in the binomial expansion the first part of the formula is

n + nx + ...

not

1 + nx + ...

as u put, u are confusing this formula with another one which u use in p2

...i think
18. (Original post by Bhaal85)
Oh no, dare I say it, your turning into AT.

Can't you go into college for some support in the morning?
You what, do you realise how badly insulted I am? How am I turning into that .... THING?

I dont want to go to college in the morning, Ill just have to listen to retards struggling to make heads or tales of every single topic in the book. I think Ill just have to wake up late, if theres too much time between waking up and the exam Ill usually get bored and want to masterbate. If I masterbate I will fail. Horrible dilemma.
19. (Original post by scottyoneill)
in the binomial expansion the first part of the formula is

n + nx + ...

not

1 + nx + ...

as u put, u are confusing this formula with another one which u use in p2

...i think
Nope... It's 1 + nx + ...
20. (Original post by shift3)
Nope... It's 1 + nx + ...
tis true.. it's 1 + nx + blah blah

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: June 9, 2004
Today on TSR

### Congratulations to Harry and Meghan!

But did you bother to watch?

### What do you actually do at University?

Poll

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE