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which is quite disappointing really because language evolves very quickly even over human lifespans society will continue to change and the law must address these changes, not ignore them or attempt to suppress them.

ok so to stipulate your ideas: you believe that language is inflexible. that summarises your entire position on this.

that's not even close

aye lmao apparently people think (1-\sqrt{h})^2 = 1-h rip there goes ur marks also a lot of people forgot that √k = 1-√h their area function was (1-√h)² instead of (1-√h)⁴ there goes another mark bless u janzen choi for being the one person who got my spiky volume correct <3 another kid...

"it's challenging and scary and complicated, therefore we shouldn't do it" hm, wonder where I've heard THAT before.

Ok firstly, scientists have figured out how to create opposite gendered zygotes for both sexes so that argument is no longer valid.

can confirm that it isn't :)

legit only one person got the spiky volumes question correct I feel attacked

Re: Extracurricular Integration Marathon Beta Function manipulation is pretty trivial....

Re: Extracurricular Integration Marathon anyway let's bump this one, been sitting here for a very long time. the first one is a trivial exercise in contour integration/special functions manipulation the second is a matter of a straightforward substitution

Re: Extracurricular Integration Marathon The function to consider is: \log{\left( \tan{\tfrac{x}{4}} \right)} \equiv \frac{2}{\pi}\sum_{k=1}^\infty \frac{(2\mathcal{H}_{2k} - \mathcal{H}_k)\sin{kx}}{k} Then by Parseval's Theorem: \frac{1}{\pi} \int_{-\pi}^\pi \log^2{\left(...

The answer to this integral is: \frac{\pi a}{\sqrt{2} \sqrt{1+\sqrt{1+a^2}}} Some methods to try: DUTIS, Contour Integration

Re: Extracurricular Integration Marathon By inspection, the answer to the first integral is: (-1)^n \zeta{(n+1)} \Gamma{(n)} Similarly, the answer to the second integral is: (-1)^{n-1} \eta{(n+1)} \Gamma{(n)} These come from the integral definitions of the Riemann Zeta and...

Observe that the term \frac{\sin{x}}{a^2+1-2a\cos{x}} \equiv \Im{\left(\frac{1}{a-e^{-ix}}\right)} This observation leads us to the immediate conclusion: Cases must be taken for the sign of log(a²) For log(a²)<0: \int_0^{2\pi} \frac{x^2 \sin{x}}{1-2a\cos{x}+a^2}\text{d}x =...

Re: HSC 2017 MX2 Marathon ADVANCED Affine Transformation Bash :P

Re: HSC 2017 MX2 Marathon ADVANCED So I believe the answer is 2tan⁻¹(b) - π/2 or something like that. Probably should have an absolute around that since we're dealing with absolute angle measures.

Re: HSC 2017 MX2 Marathon ADVANCED what do you mean by that? is the circle not allowed to have arbitrary radius? also I know what my mistake was, inverse trig simplification is a pain :P

It is not even close to doing anything with that. Homosexual couples do not have the same legal socio-economic protections as straight couples. That is the relevant issue at hand here. Not religious customs. And if you wish to drag religious customs back into this, I suggest you look into the...

Re: HSC 2017 MX2 Marathon ADVANCED WLOG, assuming a=1, the maximum angle is 2cot⁻¹(b) If this is correct, I will post the working out later, I'm a bit all over the place atm.

Re: HSC 2018 MX2 Integration Marathon \int \sqrt{\tan{x}} \, \text{d}x \equiv \frac{1}{2} \int (\sqrt{\tan{x}}-\sqrt{\cot{x}}) + (\sqrt{\tan{x}}+\sqrt{\cot{x}}) \, \text{d}x The rest is trivial.