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Eshay adlays ilchays

So jelly

GL in your test and enjoy ur hollies m8!

Re: HSC 2012 Marathon :) x=-2+1/2 do you mean -2.5?

Re: HSC 2012 Marathon :) because e^lnx = x thats why hahah

Re: HSC 2012 Marathon :) integral e^lnx = f'(x)/f(x) = 1/x woops

Re: HSC 2012 Marathon :) isnt it f'(x)/f(x) ?

Re: HSC 2012 Marathon :) Just looking at that wouldnt you make u = x^5 + 5x^2-5x ?

Re: HSC 2012 Marathon :) It's v^2=n^2(a^2-x^2)

Re: HSC 2012 Marathon :) I've always done it like that :S

Re: HSC 2012 Marathon :) <a href="http://www.codecogs.com/eqnedit.php?latex=\int \frac{x^4}{1@plus;x^{10}} \textup{dx} ~\\ Let ~u = x^5 \\ \frac{du}{dx}= 5x^4 \\dx=\frac{du}{5x^4} \\ \frac{1}{5}\int \frac{1}{1@plus;u^2}~ \textup{dx} = \frac{1}{5}\tan^{-1}u @plus; \textup{C} , ~since~ u ~= x^5...

Re: HSC 2012 Marathon :) Alright thanks,

Re: HSC 2012 Marathon :) x^10?

Re: HSC 2012 Marathon :) Requesting some hard integration questions

Re: HSC 2012 Marathon :) amplitute is 5 .

Re: HSC 2012 Marathon :) Oh i see what u did still turns out to be correct..

Re: HSC 2012 Marathon :) <a href="http://www.codecogs.com/eqnedit.php?latex=\ln (1 @plus; \frac{1}{x}) = 3t @plus; \ln(\frac{1}{2})" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\ln (1 + \frac{1}{x}) = 3t + \ln(\frac{1}{2})" title="\ln (1 + \frac{1}{x}) = 3t + \ln(\frac{1}{2})"...

Re: HSC 2012 Marathon :) Check your working for 3 there's a mistake

Find points of intersection which is (1,1) and (-1,-1) If you expand this and simplify (x-μ)^2 + (y+μ)^2 = 2(1+μ^2). You will get x^2 + y^2 = 2

Re: HSC 2012 Marathon :) For Sy, <a href="http://www.codecogs.com/eqnedit.php?latex=\textup{A ~particle~ is~ moving ~so~ that} ~\ddot{x}=18x^3@plus;27x^2@plus;9x.~ \textup{Initially ~x=-2~ and~ velocity, ~v~ is ~-6}.\\\\1. ~\textup{Show ~that~} v^2=9x^2(1@plus;x)^2 \\ 2. ~\textup{Hence, or...