Edexcel C3,C4 June 2013 Thread

lol yhyh sure
lol
Yup done, i'v been focusing on fp2 and fp3 and s3, just need to average 90 in those three to get an A*
You??

Yeah, I'm sure you can rewrite what you've got as a combination of trigs? Then all you'll have to do is IBP and you'll get the correct answer

nope im lost now lool would i be getting the right idea if i said $sin^22x \times sec2x$ and then do by parts on that?

nope im lost now lool would i be getting the right idea if i said $sin^22x \times sec2x$ and then do by parts on that?

Okay can't it be expressed as tan(2x)*sin(2x) ? Can you now integrate that by parts?

Also if you didn't want to use by parts, you could continue in this way:

O M G ! that is what i done in the beginning and i hesitated and said no to myself damn i need to believe in myself sometimes argh !! thank you anyways

Yeah, that's what TSR has given me, plus I quite enjoy trying out all these different questions that only require the A level knowledge as it's great practice for anything that might come up in the exam as it's probably a lot harder and would never be seen in a C4 exam :-/ I'd rather have questions like this in my exams as it would be more interesting

Oh yeah! I never even thought about that approach and it would definitely make it easier and less room for mistake too, thanks for that

yeh i know what you mean !! but then again if you didn't realise what to do just like me, you've lost the marks, but im happy im being challenged, allowed me to think outside the box

btw i found this pretty useful to remember the identities http://www.sosmath.com/trig/douangl/douangl.html

anyway thanks for that question phew lool back to physicssss

Does somebody know how to work out the t values:





???

Hmm I'm not quite sure how you got it to simplify into that :-/ What can the denominator be expressed as? Then is there something you know that can simplify it? Once you've done that is there anyway you could simplify the fraction? Let me know if you're still stuck

Edit: It looks like you're missing a function in your numerator, other than that's its all going the correct way so far

$\displaystyle \int \dfrac{4tan^2x}{1-tan^2x}$ = $\displaystyle \int \dfrac{4 (\frac{sin^2x}{cos^2x})}{1 - \frac{sin^2x}{cos^2x}}$ and then i cancelled out $\frac{sin^2x}{cos^2x}$ so it becomes $\displaystyle \int \frac{4}{1-tan^2x}$. Is this correct?

Ah okay, I understand where you've got that from but the denominator is 1-tan^4(x) not 1-tan^2(x) Try it again now?

$\displaystyle \int \dfrac{4tan^2x}{1-tan^4x}$ = $\displaystyle \int \dfrac{4 (\frac{sin^2x}{cos^2x})}{1 - \frac{sin^4x}{cos^4x}}$ and then i cancelled out $\frac{sin^2x}{cos^2x}$ so it becomes $\displaystyle \int \frac{4}{1-tan^2x}$. Is this correct?

Is anybody else hoping for a hard paper? I reckon I'll be fine on a hard paper because I know my maths well, I just make small mistakes sometimes. If it's a hard paper, 1 or 2 marks won't hurt too much.