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# Edexcel C3,C4 June 2013 Thread watch

1. (Original post by Rayquaza)
Yup.
did you use substitution method for the 1/(1-y^2) ?
2. (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.
3. (Original post by usycool1)

You just need to solve the differential equation.
Hi

how would you integrate

question 6)a)

http://www.school-portal.co.uk/Group...urceId=3992058

I tried to use the Sin(A+B) formula but it looks really confusing and horrible to work wit
4. (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?
5. (Original post by otrivine)
Hi

how would you integrate

question 6)a)

http://www.school-portal.co.uk/Group...urceId=3992058

I tried to use the Sin(A+B) formula but it looks really confusing and horrible to work wit
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
6. (Original post by Rayquaza)
Attachment 227150
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?
7. (Original post by usycool1)

You just need to solve the differential equation.
Yeah, that one. I kind of forgot all about differential equations! Thanks!
8. (Original post by otrivine)
Hi

how would you integrate

question 6)a)

http://www.school-portal.co.uk/Group...urceId=3992058

I tried to use the Sin(A+B) formula but it looks really confusing and horrible to work wit
Factor formulae?
9. (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

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
10. (Original post by otrivine)
Hi

how would you integrate

question 6)a)

http://www.school-portal.co.uk/Group...urceId=3992058

I tried to use the Sin(A+B) formula but it looks really confusing and horrible to work wit
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.
11. (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

So we will never be asked this question?
12. (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?
13. (Original post by usycool1)
Factor formulae?
can you show me please how to start of with
14. (Original post by Rayquaza)
So as my partial fractions I got 1/(1-y) and 1/(y+1)
Correct?
yepp and now integrate
15. (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
16. (Original post by Rayquaza)
Attachment 227145
Question 4e)

4e)

Attachment 227146

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
17. how would i integrate this
18. (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 Ignore me I've seen what you've done already
19. (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?
20. (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

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