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I think there is a mistake here, when you do: I_{2n-1} = \frac{\pi^2}{4} + \sum_{k=1}^{2n-1} \frac{(-1)^{2k-1} - 1}{(2k-1)^2} The 'summand' value isn't correct, it should be: I_{2n-1} = \frac{\pi^2}{4} + \sum_{k=1}^{2n-1} \frac{(-1)^k - 1}{k^2} Then, when you sum the series out, all the...

Is there any demand for Q14 solutions?, I can type it up quickly if someone would like it

So we get \int_{-e + e^3}^{6} \frac{du}{u} = \ln(6) - \ln(-e + e^3} = \ln 6 - \ln(e(e^2-1)) =\ln 6 - \ln e - \ln(e^2-1) = \ln \frac{6}{e^2-1} - 1 Just using log properties here

One doesn't really need to do substitution, you just need to recognise \int \frac{f'(x)}{f(x)} \ dx = \ln |f(x)| + c \int_{-e}^{-2} \frac{1-3x^2}{x-x^3} \ dx = \ln |x - x^3|_{-e}^{-2} This is a 2U concept - However if you are intent on using a substituion u = x-x^3 To get the limit...

Re: HSC 2013 4U Marathon Well the HSC method would be very similar to it, in your method, just change every e^(iy) into cos y + i sin y, and it will still work out with De Moivere's

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haahhaahahhaahahaahahahahah idiot

i can accept diamond

you want him to be egyptian sorry persian/arab you know there is something called wikipedia right

that was a horrible joke mali please never make a maths joke again 2U scum

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Yes yes yes yes yes

ibn sina

TRY AND DECODE THE SYSTEM, SEE IM NOT USING ANY LOWERCASE, HOW AM I DOING IT caps

im sorry but you just went 100% retard

get rid of your avatar i agree with mali that is a horrible sight

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ahahahahahaabaahah fail mali WHAT A FAIL YOU CANT USE CAPS

MY NAME IS SY123 HEAR ME ROAR (caps)