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Re: MX2 Integration Marathon Yeah if you intuitively can see it's xln(lnx) then you're lucky, but what IBP did you use on ln(lnx)? That seems a lot faster than my method.

cos(-225) = cos(360-225) = cos(135) = -cos(45) = -1/sqrt(2) sin(315) = sin(360-45) = -sin(45) = -1/sqrt(2)

i realised that evaluating tan(pi/18) is way out of the scope of the extension course but seriously, i thought the denominators were in the brackets. This makes such a huge difference.....

Re: HSC 2014 4U Marathon - Advanced Level probs made a mistake expanding, no way i'm going to look where hahaha

Re: HSC 2014 4U Marathon - Advanced Level wow. i can't believe i didn't see that...

alright so part i was actually pretty tough \tan{\frac{A}{5}} = k \\ A = 5\arctan{k} \\ B = 6\arctan{k} \\ C = 7\arctan{k} \\ A+B+C = (5+6+7)\arctan{k} \\ \text{Now since A B and C are the angles of a triangle, } A+B+C = \pi \\ k = \tan{\frac{\pi}{18}} \\ \text{If you evaluate that, you WILL...

Re: HSC 2014 4U Marathon - Advanced Level By subbing x = sqrt(x) you get a polynomial with roots r1^2, r2^2... etc. By expanding Q(r1)*Q(r2)*...*Q(r5). You get an expansion in terms of r1^2 r2^2 etc. This is in terms of the coefficients of the new polynomial. But I dont have paper here so I...

well not if they average like 65%. it's a passing grade but not great by any means.

Hey everyone! Subs: in sig School rank: top 30?

Rumble is 100% right, its just a mentality. Do past hsc questions and just be confident in your own ability and you'll get through it no problem. If you stress yourself out thinking you aren't good enough then you're only making it harder for yourself. Just keep calm and math :P

Re: MX2 Integration Marathon Yeah I realised that after but am yet to solve this problem, will have a crack at it tomorrow morning at school :) EDIT: ok i think i've got it now! \text{RTP:} \; I = \int_a^{a+T}{f(x)dx} = \int_0^T{f(x)dx} \: \: , \: \: a \in \mathbb{R} \\ \equiv...

Re: MX2 Integration Marathon ok so what we are trying to prove is that \int_0^T{f(x)dx} = \int_{(k-1)T}^{kT}{f(x)dx} i'm not sure how valid my method is for this but RHS = I = \int_{(k-1)T}^{kT}{f(x)dx} \\ \\ u = x - (k-1)T \\ du = dx \\ u_1 = T \\ u_2 = 0 \\ I = \int_0^T{f(u+(k-1)T)du}...

Re: HSC 2014 4U Marathon good question! took me a couple of minutes to wrap my head around what the solid looked like but then i realised it said the cross sections were perpendicular not parallel, and i felt like a fool :( loving the volumes, it's definitely my favourite topic so far :) EDIT...

take it slowly and think it through logically, and most importantly of all, do not over complicate it. the majority of the questions are actually extremely simple.

I'm with rumble on this. Analysing literature is one of the most painstaking and useless tasks that has ever been required of me. I would have no problem with English being compulsory if it achieved it's purpose (teaching students to write and communicate ideas), but it doesn't. I can write a...

Re: MX2 Integration Marathon \int{\left (\frac {1}{lnx} + ln (ln (x))\right ) dx

Re: MX2 Integration Marathon \text {Prove, without the use of complex numbers that} \\ \\ \int_0^{2\pi} {\ln{\left (\frac {\left (1+sinx \right )^{1+cosx}}{1+cosx}\right )}}dx = 0

anyone who starts the course that far back is honestly retarded. you gain absolutely nothing. i'm doing the course currently, and there are a few kids who started it at tutoring at the start of year 11. Needless to say, none of them have even broken the top 15. The majority of the top of our...

Re: MX2 Integration Marathon Ah right, that makes a lot of sense. I got it to that stage but I always forget that partial fractions is an option with trig :P

Re: MX2 Integration Marathon not sure if i got this right, but i got \\ \frac{1}{6}\left({-\sqrt{2}\tan^{-1}{\left (\frac{1}{2}\right )}} - \ln{\left | \frac{\sqrt{2} + 2}{2}\right |} + \ln{\left | \frac{\sqrt{2} - 2}{2}\right |} \right ) i substituted u = sinx then did partial...