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Edexcel C3,C4 June 2013 Thread

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Original post by Rayquaza
Yup.


did you use substitution method for the 1/(1-y^2) ?
Reply 3841
Original post by otrivine
did you use substitution method for the 1/(1-y^2) ?


No...
Think that's where I went wrong. I'll try it now.
Original post by usycool1
The one about volumes?

You just need to solve the differential equation. :smile:


Hi :smile:

how would you integrate

question 6)a)

http://www.school-portal.co.uk/GroupDownloadFile.asp?GroupId=1152243&ResourceId=3992058

I tried to use the Sin(A+B) formula but it looks really confusing and horrible to work wit:frown:
Original post by Rayquaza
No...
Think that's where I went wrong. I'll try it now.


Surely you don't need a substitution just expand the denominator as its the difference of two squares, so can't you separate by partial fractions and then integrate each term separately?
Original post by otrivine
Hi :smile:

how would you integrate

question 6)a)

http://www.school-portal.co.uk/GroupDownloadFile.asp?GroupId=1152243&ResourceId=3992058

I tried to use the Sin(A+B) formula but it looks really confusing and horrible to work wit:frown:


Check your formula booklet and you should be able to see something that looks similar in it. If not just ask and I'll help you out :smile:
Original post by Rayquaza
ImageUploadedByStudent Room1371376909.444296.jpg
This is my working. Try to make sense of it, its a bit all over the place.


Posted from TSR Mobile


wouldn't you do partial fractions rather than reverse chain rule?
Original post by usycool1
The one about volumes?

You just need to solve the differential equation. :smile:


Yeah, that one. I kind of forgot all about differential equations! :colondollar: Thanks!
Original post by otrivine
Hi :smile:

how would you integrate

question 6)a)

http://www.school-portal.co.uk/GroupDownloadFile.asp?GroupId=1152243&ResourceId=3992058

I tried to use the Sin(A+B) formula but it looks really confusing and horrible to work wit:frown:


Factor formulae? :smile:
Original post by MathsNerd1
Surely you don't need a substitution just expand the denominator as its the difference of two squares, so can't you separate by partial fractions and then integrate each term separately?


Hi :smile:

so do you mean
use


oh yes, true,
I can use partial fractions and integrate separately, cause I was trying the sub method but looks confusing to integrate
Original post by otrivine
Hi :smile:

how would you integrate

question 6)a)

http://www.school-portal.co.uk/GroupDownloadFile.asp?GroupId=1152243&ResourceId=3992058

I tried to use the Sin(A+B) formula but it looks really confusing and horrible to work wit:frown:


Although it's on the Solomon papers, it's not on the Edexcel spec, even though the factor formulae are given in the formula book.
Original post by tiny hobbit
Although it's on the Solomon papers, it's not on the Edexcel spec, even though the factor formulae are given in the formula book.


Hi :smile:

So we will never be asked this question?
Reply 3851
Original post by masryboy94
wouldn't you do partial fractions rather than reverse chain rule?



Original post by MathsNerd1
Surely you don't need a substitution just expand the denominator as its the difference of two squares, so can't you separate by partial fractions and then integrate each term separately?


So as my partial fractions I got 1/(1-y) and 1/(y+1)
Correct?
Original post by usycool1
Factor formulae? :smile:


can you show me please how to start of with :smile:
Original post by Rayquaza
So as my partial fractions I got 1/(1-y) and 1/(y+1)
Correct?


yepp and now integrate
Original post by Rayquaza
So as my partial fractions I got 1/(1-y) and 1/(y+1)
Correct?


That's almost correct but I got a 1/2(1+y) + 1/2(1-y) can you see where you went wrong? Ignore this as this would only be for if you're separating 1/(1-y^2) and not the 2/(1-y^2) that you have done :smile:
(edited 10 years ago)
how would i integrate this sec33x \int sec^3 3x
Reply 3856
Original post by Rayquaza
ImageUploadedByStudent Room1371375876.394132.jpg
Question 4e)


Answer:
4e)

ImageUploadedByStudent Room1371375946.088925.jpg

Getting a weird decimal for '+c'
Help?



Posted from TSR Mobile


you method just before integration is correct: split the fraction of 2/(1-y^2) up into partial fractions. you should get 1/(1-y) and 1/(1+y). Integrate (1/(1-y) + 1/(1+y)) dy = integration 1/(1+x) dx After integration you will be left with ln(1+y) - ln(1-y) = ln(1+x) + c sub into the above x=5 and y=0.5 rearrange to get c=ln(0.5) rewrite the whole equation as; ln(1+y) - ln(1-y) = ln(1+x) + ln(0.5) now put it all together; ln((1+y)/(1-y)) = ln ((1+x)/2) sorry it s bit messy!!!! where ever I wrote 0.5 think of it as 1/2
Original post by masryboy94
yepp and now integrate


Are you sure about that? Try expanding your partial fractions and you'll notice that it won't equal 1/(1-Y^2) unless I'm just being stupid :redface: Ignore me I've seen what you've done already :smile:
(edited 10 years ago)
Reply 3858
Original post by tiny hobbit
Although it's on the Solomon papers, it's not on the Edexcel spec, even though the factor formulae are given in the formula book.

It isn't on the spec? then why are there questions on it in the edexcel text book?
Original post by MathsNerd1
Are you sure about that? Try expanding your partial fractions and you'll notice that it won't equal 1/(1-Y^2) unless I'm just being stupid :redface:


Your partial fractions are correct :tongue:

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