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# P5 integration problem watch

1. (Original post by Aitch)
This is what I've got (in paper form), and I still can't find the above! I can find them in the AQA version OK. Perhaps I'm going mad!

http://www.edexcel.org.uk/VirtualCon...cification.pdf

I've learned them anyway...

Aitch
But they are in my old (Green) Edexcel Formula Book!

Aitch
2. Ah right. I'm looking at the 'old' syllabus one. I assume I'm gonna get the old syllabus booklet, since I sat P1-P4, and so it would follow that I'd be sitting P5 and P6 rather than FP2 and FP2 (if you know what I mean).
Strange they don't put it in the new one...
3. (Original post by mockel)
Ah right. I'm looking at the 'old' syllabus one. I assume I'm gonna get the old syllabus booklet, since I sat P1-P4, and so it would follow that I'd be sitting P5 and P6 rather than FP2 and FP2 (if you know what I mean).
Strange they don't put it in the new one...
Could be another error. The trig identities are now listed under C3 in the new booklet, so someone has probably removed the tan(t/2) identities from there, and forgotten to paste them in elsewhere.

Aitch
4. integrate 1/(1+x^4)
i use subtitution of x=tan u at the end i got 1/2 { lnA[sec(2 arctanx) + 2x/(1-x^2)]}
seems totally different from wat u guys got
ah this ques did come up in STEP III last year, they put the limit in and there was some hint and work at the beginning...
btw nyone know how to integrate e^sinx ?
5. Can anyone help me with this? I'd be v grateful, thanks.

Integrate: x+2/sqrt[x^2 + 6x + 4]
6. split it up so u get (2x + 6)/2sqrt(x^2 + 6x + 4) - 1/2sqrt(x^2 + 6x + 4)
7. (Original post by riks2004)
split it up so u get (2x + 6)/2sqrt(x^2 + 6x + 4) - 1/2sqrt(x^2 + 6x + 4)

thanks, i'll give it a go.

sorry, doesn't work. If you put your split fractions back together u don't get what u started with. u'll end up with:

x+2.5/sqrt[x^2 + 6x +4] rather than x+2 as the numerator.

i tried splitting it like:

x/sqrt[x^2 + 6x + 4] + 2/sqrt[(x+3)^2 - 5]

for the second part u can take the 2 outside the integral and use the standard form of 1/sqrt[x^2 - a^2]. any ideas on the integral of the first part tho? anyone?

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