Recent content by omegadot

$\noindent Here is a real method, though it is nowhere near as slick as that given by seanieg89.$ $\noindent We begin by splitting up the interval of integration as follows:$ \int^{2\pi}_0 \cos^{2n} \theta \, d\theta = \int^{\frac{\pi}{2}}_0 \cos^{2n} \theta \, d\theta +...

$\noindent So do I, so if you can give me a low tech approach that makes use of real methods only for the integral$ $$\int_{-\infty}^{\infty}\frac{dx}{(e^{x}-x)^{2}+{\pi}^{2}}=\frac{1}{1+\Omega},$$ $\noindent where $\Omega$ is the omega constant, I'd be really impressed.$

Re: HSC 2018 MX2 Integration Marathon $\noindent Now, so as not to scare off the children too soon, how able something a little easier?$ $\noindent Find, without using a trigonometric substitution, $ \int \frac{dx}{x^2 \sqrt{x^2 + 1}}.

Re: HSC 2018 MX2 Integration Marathon $\noindent Note, the (improper) integral only converges for $n > 1. $\noindent The simplest way to solve this integral is to use a hyperbolic substitution of $x = \sinh t$, though as hyperbolic functions are not considered part of the MX2 syllabus I...

$\noindent Evaluate $\int_0^1 \frac{\ln x \ln (x + 1)}{x (x + 1)} \, dx.

Re: HSC 2018 MX2 Integration Marathon $\noindent Evaluate$ \int_{-1}^1 \frac{e^{2x} + 1 - (x + 1)(e^x + e^{-x})}{x(e^x -1)} \, dx.

Re: Extracurricular Integration Marathon Beta function, second derivative with respect to its parameter to get rid of the log squared term, beta function reflection formula is then related to the digamma function (and of course its second derivatives will be needed along the way), then a...

Re: Extracurricular Integration Marathon $\noindent I will confine myself to the real domain. Not too sure about the `trivial exercise in special functions manipulation' but I guess `trivial' is in the eye of the beholder.$ $\noindent In solving this problem the special functions I am...

Re: Extracurricular Integration Marathon $\noindent I will help fill in the details for (a). (b) is done in a similar fashion.$ $\noindent The two important pieces of information needed in order to solve the first integral is the definition of the zeta function $\zeta(s)$ and the gamma...

In addition to those texts listed by seanieg89 I would add: 1. The Soviet text Problems in Mathematical Analysis edited by Boris Demidovich has many problems using the various techniques of integration to solve. A few of the techniques Demidovich gives are not usually found elsewhere. 2...

Re: HSC 2017 MX2 Integration Marathon $\noindent There is quite a nice way to do the third one. We begin by noting that$ \begin{align*}\sin 2x &= 2 \sin x \cos x = 1 + 2 \sin x \cos x - 1 = (\cos^2 x + \sin^2 x) + 2\sin x \cos x - 1 = (\cos x + \sin x)^2 - 1.\end{align*} $\noindent So...

Re: HSC 2017 MX2 Integration Marathon $\noindent The second one can also be found using the reverse product rule. Begin by noting that$ \begin{align*}\left (\frac{1 - x}{1 + x^2} \right )^2 &= \frac{(1 - x)^2}{(1 + x^2)^2} = \frac{1 + x^2}{(1 + x^2)^2} - \frac{2x}{(1 + x^2)^2} =...

I don't think NESA really care. As long as they can say they are making changes that modernise their syllabuses, which follow current "best practice", and grab a headline or two along the way is all that counts.

Considering NESA cannot even handle the current change effectively, how on earth are they going to manage changes in the syllabuses every five years?

Re: Extracurricular Integration Marathon $\noindent If there is still interest in summing a few infinite series you might like to try these two:$ $\noindent (a) $ \sum^\infty_{n = 1} \frac{1}{n(n + 1)(2n + 1)} \quad $(b) $\sum^\infty_{n = 1} \frac{(-1)^{n + 1}}{n(n + 1)(2n + 1)}.