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Re: MX2 Integration Marathon Yep that is correct so far, the rest is relatively easy.

PMed.

Re: HSC 2013 4U Marathon HINT: Integration

Re: HSC 2013 4U Marathon Yep that's true, edited.

Re: MX2 Integration Marathon I(n,m) = \int \frac{1}{x^n(x-a)^m} \ dx I(n,m) = \frac{-1}{(n-1)x^{n-1}(x-a)^m} - \frac{m}{n-1} I(n-1,m+1) Set m=1, keep applying recurrence formula, then simplify: I(n,1) = \sum_{r=1}^{n-1} \frac{(-1)^r (n-r-1)!}{(n-1)! x^{n-r}(x-a)^r} +...

Re: MX2 Integration Marathon Are you sure that is integrable in terms of elementary function? I_n = \int_{0}^{k} (k^2-x^2)^n \ dx $Prove that$ I_n = \frac{2(2k)^{2n}}{(n+1) \binom{2n+2}{n+1}}

Re: HSC 2013 4U Marathon Good job. Ah yes but I've seen 3U questions (like in 2012 Independent) where they simply replace the m with 2. Because that is easier to integrate at 3U level (substitution). But looking at the LHS 'reminds' you with something to do with integration, which may lead...

Of course, understanding Physics is a lot more than knowing how to calculate stuff, but at least for me, knowing what rates of change are and calculus, really helps in my understanding of Physics concepts, even in Motors. I think the hardest part of Motors may be Back EMF, but I think that's it...

y = \cos^{-1}(\sin x) \ \ \ -\pi \leq x \leq \pi y '= \frac{-1}{\sqrt{1-\sin^2 x}} \cdot \cos x y' = \frac{- \cos x}{|\cos x|} y' = -1 \ \ $for$ \ \ \cos x \geq 0 y'= 1 \ \ $for$ \ \ \cos x \leq 0 This is because the cos x on top cancels out the cos x at the bottom, but for different...

If you do a sufficient level of maths Physics 'understanding' becomes quite easy because the calculations and such are all really simple, from then on you need to know all about bs things like how rockets work and appliances etc etc. which is a lot easier compared to Chemistry (at least for me)...

Be about the top 4000 of students in the state But seriously you can't really determine that without knowing your school rank.

Re: HSC 2013 4U Marathon Well I didn't, you may find a way using it, this question is more about argument than algebraic skill (like my series question just now)

Re: HSC 2013 4U Marathon Definitely not :P Though I am curious to see how you would do so. EDIT: Yes that is the method that I had in mind for this question (though I didn't use the Beta function specifically), you need to 'think up of' a function such that you integrate you get the LHS, the...

Re: MX2 Integration Marathon Alright nice.

Re: MX2 Integration Marathon Ok here is my attempt at proving the above equality. We can establish that: \int_0^{\pi /4} \ln(\sin(2x)) \ dx = \frac{1}{2} \int_{0}^{\pi /2} \ln(\sin(x)) \ dx By simply using the substitution 2x = u And since sin(2x) is symmetrical about pi/4 from x=0 to...

Re: MX2 Integration Marathon I am guessing I need to multiply 1 in a clever way, no? I need to find a way to prove that: \int_0^{\pi/2} \ln(\sin(2x)) = \int_0^{\pi /2} \ln(\sin(x)) \ dx I am considering some sort of periodicity argument but I don't think that might work.

Re: MX2 Integration Marathon For the cot one, is it a really obscure substitution of the difficulty of like x = (1-u)/(1+u)?

Re: HSC 2013 4U Marathon $Prove that$ \sum_{r=0}^{n} \frac{(-1)^{r}}{m+r+1} \binom{n}{r} = \frac{m! n!}{(m+n+1)!} (there is a reason I asked this in this 4U thread)

Re: MX2 Integration Marathon Yep, good job. ======= By simplifying the integral and using inverse trig rules, we end up with -pi/2 x + C ====== \int_0^a \frac{dx}{x+\sqrt{x^2-a^2}}

Re: MX2 Integration Marathon \int_0^{\pi/2} \frac{dx}{1+ \tan^{\pi}(x)}