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Reply 20
and for the second one ive got dx = 1/2 u, so i know to swap the dx with this 1/2 u but just dont understand what to do with the (4x-8)(2x-5)^7 part, i realise the (2x-5)^7 becomes u^7 but what about the 4x-8? thanks
Reply 21
i need 78 in this and m2 to get an A. i think its doable but my confidence was destroyed by c3 in january, i seem to do better at c4 and m2 tho, even if my m2 teacher is terrible
Reply 22
jkkk32
a = 4i + 3j - 2k and b = -7i + 5j + 4k
is a.b like 4x-7 + 3x5 + -2x4 here?? i thought its between the direction vectors? not position vectors
and i still dont understand how to find the angle from my first answer?

No because you want the angle between OAB so you need to use the position vector that you found in the first part (remember when you had to find AB which is B-A?) so you want the angle between the lines OA and AB. You know that the position vector for OA is 4i + 3j - 2k and from (i) AB = -11i + 2j +6.
and then you apply your scalar product.

jkkk32
and for the second one ive got dx = 1/2 u, so i know to swap the dx with this 1/2 u but just dont understand what to do with the (4x-8)(2x-5)^7 part, i realise the (2x-5)^7 becomes u^7 but what about the 4x-8? thanks

Yes so you put the du/2 into the equation. Which means that you divide the whole equation by 2 (because you're multiplying it by 1/2 du). Rearrange your u equation to get x = (u+5)/2 and substitute that in (You should get (2x-4) x u^7 before you substitute it in. Then you integrate it all after simplifying it.
Reply 23
gonna need high ums for A* and kinda need it lol gl me.

just learnt how to integrate cos^2(x) and revising vectors now.

glgl hope it turns out well for everybody.... :]
Reply 24
happi_minnie
does anyone have jan 2010 paper? i've got the mark scheme but no paper :frown:


yeah i have it, pm me your email if you want it and i'll send it, happi_minnie!
Reply 25
ziigmund
gonna need high ums for A* and kinda need it lol gl me.

just learnt how to integrate cos^2(x) and revising vectors now.

glgl hope it turns out well for everybody.... :]


How do you integrate it?

Thanks in advance
Reply 26
elliot22
Does anyone know where I could find OCR Core 4 June 2005 Mark Scheme?

I can't find it anywhere online!


I've got it elliot22, pm me your email and i'll send it to you! :smile:
Cos^2x can be integrated using cos2x. But do you know why you can take 1/2 out of the integration?
Reply 28
crazy4it2007
Cos^2x can be integrated using cos2x. But do you know why you can take 1/2 out of the integration?

Why you can? Because it's a constant? That's not going to be on the exam is it? :eek3:
Damn you core 3. I was aiming for an A*, but after today's paper, its impossible now :frown:

Anyone got the core 4 mark scheme from jan 2010?
Reply 30
how do you integrate cos²2x ??
Reply 31
cos 4 x = cos ( 2x + 2x ) = 2cos^2 2x - 1

cos^ 2 x = 1/2 (1 + cos 4x) then intergrate.
Woahhh hang on, OCR Core 4 A2?????
My exam was yesterday??????!!!?????? :O
Guys...........unless its a different number sorta thing...

Its no 4754/01 is it?
not**
Reply 35
We're doing 4724/01 :troll:
Oh thats alright then! I nearly had a panic attack ! :biggrin:
I've never felt so pressured by a maths exam before... :/
rev_pain
how do you integrate cos²2x ??


As I've never seen the one with cos4x before (it looks complicated) here's how I derive it.

You have the trig identities:

sin²x+cos²x=1

and your formula for cos(2x)=cos²x-sin²x

Then you sub in sin²x from your first identity to your second which gives:

cos2x=2cos²x-1

so cos²x=1/2(cos2x+1)

(I think. Don't have a calculator to check From memory you can do the same with sin²x, except it's sin²x=1/2(sin2x-1)).

When you integrate this you can just take 1/2 out yeah.
Reply 39
MarcusTheEskimo
As I've never seen the one with cos4x before (it looks complicated) here's how I derive it.

You have the trig identities:

sin²x+cos²x=1

and your formula for cos(2x)=cos²x-sin²x

Then you sub in sin²x from your first identity to your second which gives:

cos2x=2cos²x-1

so cos²x=1/2(cos2x+1)

(I think. Don't have a calculator to check From memory you can do the same with sin²x, except it's sin²x=1/2(sin2x-1)).

When you integrate this you can just take 1/2 out yeah.


Your forgetting that its cos^ 2 2x not cos2 x

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