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Edexcel A2 C4 Mathematics June 2015 - Official Thread

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Original post by raypalmer
Hopefully vectors aint the last question, when it is its usually a mad ting


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I can't do any of c4 apart from vectors at the moment lol
too many exams in a row
Original post by Paraphilos
See my post above.


the question is untitled.JPG

If I want to solve this q. by substitution can I make cosx = u?
(edited 8 years ago)
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mb_12
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Original post by MrBowcat
So what what would you do exactly after you get cosx - cosx sin^2x ??


Integral of cosx sin^nx dx
= (1/n+1)sin^(n+1)x + c
This is something you are expected to be able to recognise in an exam.

I could do the question on paper if you like?
Original post by frozo123
I can't do any of c4 apart from vectors at the moment lol
too many exams in a row


same here! learning integration now and after that need to go through all vector questions and differential equations :frown:
Original post by frozo123
is it y=2(e^tan3x+1)?


That is not what I got from the brief calculations I just did; you should initially arrive at:

∫2yy dy=3∫π4xsec⁡2(x) dx\displaystyle\int_2^y y \ dy = 3\displaystyle\int_{\frac{\pi}{4}}^x \sec^2(x)\ dx

From there, can you see where to go?
Original post by manlikem
the question is untitled.JPG

If I want to solve this q. by substitution can I make cosx = u?


yes you can
Original post by manlikem
thanks! and if I use substitution could I make cos x = u?


That is exactly what you should do, in my opinion. At least, that is a systematic method; as one poster has stated above, there are general formulae for the integration of functions which are comprised of both sine and cosine functions but you can do it from first principles in the exam by using a suitable substitution. The general formulae for each case is derived in the same way.
Original post by sj97


took me a while to figure it out...at first I just substituted normally but the x value i got was -788 which was wayyyyy tooo big so I knew it was wrong...

its 0.1 because you see the limits they give you, its x<8/9
therefore you have to convert 7100 into 7.1 before you can solve to find x...


I can't really draw it on here...
ummm

(7100)^1/3 = (1000)^1/3 (7.1)^1/3 = 10(7.1)^1/3
then you can do 8-9x = 7.1
x = 0.1

then the rest is just standard :smile:

sorry it took me a while... I was doing part a as well :smile:
Original post by sincostanxxx
Can anyone explain question 8F from the IAL January 2014 paper (link: http://www.physicsandmathstutor.com/a-level-maths-papers/c4-edexcel/ )
Thank you ^.^

Draw it. Basically find AX, and find the general vector from X to a point on line2 (line2 general vector minus X). Find when this vector magnitude is equal to the AX magnitude you found. This should give you 2 parameter values, which you sub in to get b1 and b2.
Original post by frozo123
I got an idea for you
stand 2m away from your wall
step 1
Walk forward ( perpendicular ) until your hand bangs against the wall

after that
Stand in the exact same spot, but try touching the wall at the end of the room

which one took less less time and distance?


Loool good thing i put my hand there or i would've been knocked out for the exam tomorrow

Thanks
Original post by Paraphilos
That is not what I got from the brief calculations I just did; you should initially arrive at:

∫2yy dy=3∫π4xsec⁡2(x) dx\displaystyle\int_2^y y \ dy = 3\displaystyle\int_{\frac{\pi}{4}}^x \sec^2(x)\ dx

From there, can you see where to go?


ye I did that I and got lny=tan3x+ c
so c= ln2 +1

then I rearranged it, what have I done wrong?
edit: actually wait
is it tanx?
Original post by arsenalfc97
Loool good thing i put my hand there or i would've been knocked out for the exam tomorrow

Thanks


My teacher explained it like that but with heads banging and seems like our class remembers it lol
Original post by mb_12
Integral of cosx sin^nx dx
= (1/n+1)sin^(n+1)x + c
This is something you are expected to be able to recognise in an exam.

I could do the question on paper if you like?


Yes please, would greatly appreciate that!
Original post by Paraphilos
That is exactly what you should do, in my opinion. At least, that is a systematic method; as one poster has stated above, there are general formulae for the integration of functions which are comprised of both sine and cosine functions but you can do it from first principles in the exam by using a suitable substitution. The general formulae for each case is derived in the same way.


thanks! because on the mark scheme it says that I have to make sinx= u. But I in an exam I would have made cosx = u so I was confused. Does it matter which value you make u ?
Original post by frozo123
ye I did that I and got lny=tan3x+ c
so c= ln2 +1

then I rearranged it, what have I done wrong?
edit: actually wait
is it tanx?


First, note that the LHS consists of yy rather than 1y\dfrac{1}{y}; and yes, the integral of that function on the RHS is tan⁡(x)\tan(x) :wink:
Original post by Maham88
yes you can


thaaanks!
because in the mark scheme they made sinx= u and not cosx =u so I was confused
Original post by MrBowcat
So what what would you do exactly after you get cosx - cosx sin^2x ??


Reverse chain rule, for sin^3x. That would turn out to be sinx- 1/3sin^3x.
Original post by Paraphilos
First, note that the LHS consists of yy rather than 1y\dfrac{1}{y}; and yes, the integral of that function on the RHS is tan⁡(x)\tan(x) :wink:


omg I'm freaking out I can't integrate
so if there's a coefficient in a trig function you keep it
but if the angle is say k
you do 1/k times the coefficient and integration of the trig function right?

so is the answer
y=2(e^3tanx-3 ) ?
Original post by manlikem
thanks! because on the mark scheme it says that I have to make sinx= u. But I in an exam I would have made cosx = u so I was confused. Does it matter which value you make u ?


Well you can do it using that method but that makes less sense; in their case, they presumably want you to write, using their substitution:

∫cos⁡3(x)sin⁡(x) dx=∫(cos⁡2(x)sin⁡(x))cos⁡(x) dx=∫(1−u2)u du\displaystyle\int \cos^3(x)\sin(x) \ dx = \displaystyle\int (\cos^2(x)\sin(x))\cos(x) \ dx = \displaystyle\int (1 - u^2)u \ du

Which just seems like more work, albeit not a whole lot more.
Original post by frozo123
omg I'm freaking out I can't integrate
so if there's a coefficient in a trig function you keep it
but if the angle is say k
you do 1/k times the coefficient and integration of the trig function right?

so is the answer
y=2(e^3tanx-3 ) ?


Don't worry, it's a lot easier than you think; 33 is just a constant - forget about it somewhat. It is simply a multiplication factor of the RHS integral. The only real calculations you need to worry about are:

∫y dy\displaystyle\int y \ dy and ∫sec⁡2(x)dx\displaystyle\int \sec^2(x) dx

If there is a coefficient to the angle variable then there is going to be some change, yes. But in this case there is nothing like that. The answer is not the one you have given.

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