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Re: HSC 2017 MX2 Integration Marathon No that would be a bad idea as the Extracurricular Integration Marathon deals with integrals at a level well beyond those expected at the MX2 level.

Re: HSC 2017 MX2 Marathon Why indeed it is! As asked, the question makes for a very nice complex number question (and a bit of algebra) which is more than doable at the MX2 level.

Re: HSC 2017 MX2 Integration Marathon $\noindent And tends to be quite difficult to apply in many instances. Don't believe me? Well try it on $\int \frac{x^2 + 20}{(x \sin x + 5 \cos x)^2} \, dx.

Re: HSC 2017 MX2 Integration Marathon $\noindent Not sure why. Of course the name of the integral was not given in the test paper. From the statistics I have seen for that year [\textit{Amer. Math. Monthly}, Vol. 113, pp. 733-743 (2006)] the authors report that of the top 196 contestants...

Re: Extracurricular Integration Marathon $\noindent I didn't realise the problem was so interesting until I started playing around with it but I should have suspected this was the case given the original question as asked was first solved by G. H. Hardy. $\noindent Starting with...

Re: HSC 2017 MX2 Integration Marathon $\noindent Find $\int \sqrt{1 + \sin 2x} \, dx.

Re: HSC 2017 MX2 Integration Marathon $\noindent Here's and alternnative solution to this integral will a few interesting side notes along the way.$ $\noindent Let $u = \tan x, du = \sec^2 x \, dx = (1 + \tan^2 x) \, dx$ or $dx = \frac{du}{1 + u^2}. $\noindent For the limits, when $x = 0, u...

Re: HSC 2017 MX2 Integration Marathon $\noindent To follow on from this question. Hence evaluate $\int^{\frac{\pi}{2}}_0 \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} \, dx.

Re: HSC 2017 MX2 Integration Marathon $\noindent There is indeed a way but it is not at all obvious to begin with. We proceed by taking the following two steps. In the first case we write the numerator so the denominator appears. In the second case we write the numerator so the derivative...

Re: HSC 2017 MX2 Integration Marathon $\noindent A slightly easier one. Evaluate $\int^3_1 \frac{\ln x}{3 + x^2} \, dx.

Re: HSC 2017 MX2 Integration Marathon $\noindent This integral is hardly easy as one must handle the limits of integration with great care. From the hint we have$ \begin{align*}4 \ln (\sqrt{1 + x} + \sqrt{1 - x}) \ln (\sqrt{1 + x} - \sqrt{1 - x}) &= \left [\ln(\sqrt{1 + x} + \sqrt{1 - x})...

Re: Extracurricular Integration Marathon $\noindent Evaluate (preferably using real methods) $\int^\infty_0 \frac{\ln x}{1 + x^3} \, dx.

Re: HSC 2017 MX2 Marathon $\noindent Let $n$ be a positive integer and let $\zeta_1, \zeta_2$, and $\zeta_3$ be the roots of the equation $z^3 + 1 = 0$. Here $z \in \mathbb{C}. $\noindent If $a> 0$ show that $\frac{1}{3} \sum^3_{k = 1} \frac{1}{|\zeta_k - a|^2} = \frac{1 + a^2 + a^4}{(1 +...

$\noindent Not that I would wish this on anybody, but brute force also gives all three real solutions. Let $a = 2^x > 0$ for all $x \in \mathbb{R}. $\noindent Writing the equation as$ (a - 4)^3 + (a^2 - 2)^3 - (a^2 + a - 6)^3 = 0 $\noindent on factorising this becomes$ -3(a - 2)(a +...

$\noindent Note that $x \neq 0,\pm 1. $\noindent Let $u = \frac{x -1}{x + 1}$. Thus $x = \frac{1 + u}{1 - u}$ and the functional equation becomes$ f(u) + f \left (\frac{u - 1}{u + 1} \right ) + f\left (-\frac{1}{u} \right ) = \frac{1 + u}{1 - u} \qquad (1) $\noindent And let $u =...

$\noindent Here is one way to show $a$ and $b$ are unique. Let $u = a + b \omega$. If $a$ and $b$ are not unique, then it is possible to write this complex number as $u = \alpha + \beta \omega$. Here $a,b,\alpha, \beta \in \mathbb{R}$ while $\omega$ is the complex cube root of unity whose...

Re: Extracurricular Integration Marathon $\noindent I have a solution for this one, but given it is quite long, I will break it down into a number of parts.$ $\noindent Paradoxia - If you have a shorter or more elegant solution compared to what I am about to give, please share it with the...

Re: Extracurricular Integration Marathon $\noindent Using real methods only, evaluate $\int^\infty_0 \frac{\sqrt{x}}{(1 + x)^2} \ln x \, dx.

$\noindent Correct. Or from the second part, when $k = 1$ it is well known that the equation $|z - z_1| = |z - z_2|$ gives the equation of a line for the perpendicular bisector joining the points $z_1$ and $z_2$. Since in our proof above we set $\alpha = 1 - k^2$ this is equal to zero when $k =...

Re: Extracurricular Integration Marathon $\noindent In this question the following result will be used:$ \int^\infty_0 \frac{\sin ax}{x} \, dx = \left \{\begin{align*}\frac{\pi}{2}, \quad a > 0\\[2ex]-\frac{\pi}{2}, \quad a< 0.\end{align*}\right. $\noindent Now$...