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Re: MX2 2016 Integration Marathon $\noindent Mmm...improper integrals. Not sure if they are still in the syllabus either. I know they were asked way back when, and I know a section on them can still be found in Coroneos' 4 Unit book. But it is such a nice question, so here we go...

Re: MX2 2016 Integration Marathon Of course developing and using the hyperbolic functions indirectly in any given question could be expected, but employing a hyperbolic substitution straight off the bat in order to find an integral is what I was really asking about.

Re: MX2 2016 Integration Marathon I didn't think the hyperbolic functions and their inverses were part of the standard MX2 course. Am I mistaken?

Re: MX2 2016 Integration Marathon $\noindent Going back to Paradoxica's 4th integral, once we get to$\\I = \int \frac{\sqrt{u^2 + 2u + 2}}{u + 2} \, du = \int \frac{\sqrt{v^2 + 1}}{v + 1} \, dv\\$the following substitution (a so-called Euler substitution) can be used:$ \sqrt{v^2 + 1} = v + t...

Re: HSC 2016 4U Marathon $\noindent And here is what it looks like on an Argand diagram. The red curve corresponds to $\left |z + \frac{2}{z} \right | = 2$, the blue curve to the circle with minimum radius $\sqrt{3} - 1$, while the green curve to the circle with maximum radius of $\sqrt{3} + 1.

Re: HSC 2016 4U Marathon $\noindent Geometrically $|z| = k$ where $k > 0$ is a circle in the Argand plane. So one is required to find the circle with the greatest radius subject to the constraint $\left |z + \frac{2}{z} \right | = 2$ (which is also just some curve, or be it rather...

Re: MX2 2016 Integration Marathon $\noindent I am sorry Paradoxia but I don't quite have a simple three-liner solution to this one. Your solution is none-the-less very impressive! To address ze-'s concern about the level of difficult of questions asked here, let me rewrite my question in a form...

Re: MX2 2016 Integration Marathon $\noindent Let $x = u^2$, so that $dx = 2u \, du.$ Thus\\\begin{align*}\int \sqrt{\frac{x}{1- x}} \, dx &= \int \frac{2u^2}{\sqrt{1 - u^2}} \, du\\&= -2 \int \frac{(1 - u^2) - 1}{\sqrt{1 - u^2}} \, du\\&= 2 \int \frac{du}{\sqrt{1 - u^2}} - 2 \int \sqrt{1 -...

$\noindent Something a MX2 student may find either interesting or intriguing is the closed-form expression for the following infinite exponential tetration:$ h(x) = x^{x^{x^{.^{.^{.}}}}} = \frac{\mbox{W}_0 (-\ln x)}{-\ln x}, \,\, \mbox{for} \,\, e^{-e} \leqslant x \leqslant e^{1/e}. $\noindent...

Re: HSC 2016 4U Marathon $\noindent Here is an algebraic approach which makes use of a number of inequalities stemming from the triangle inequality. $\noindent For the \textit{mimimum} value, using $ |z_1 - z_2| \geqslant \Big{|}|z_1| - |z_2| \Big{|}. $\\Clearly $|z_1 - z_2| \geqslant |z_1|...

Re: HSC 2016 4U Marathon $\noindent \textbf{Next Question} $\noindent For all complex numbers $z,w$ satisfying $|z| = 12$ and $|w - 3 - 4i| = 5$, find the minimum and maximum value of $|z - w|.

Re: HSC 2016 4U Marathon $\noindent To show $\sqrt[n]{a}, \sqrt[n]{b},$ and $\sqrt[n]{c}$ forms a triangle we will show that $\sqrt[n]{a}, \sqrt[n]{b},$ and $\sqrt[n]{c}$ satisfy the triangle inequality rule given that $a, b,$ and $c$ are positive and satisfy the triangle inequality rule...

So for the best text which is not "publicly" available, I am guessing you are referring to the text from Sydney Grammar?

Re: HSC 2016 4U Marathon $\noindent \textbf{Next Question} $\noindent If $v,w,z \in \mathbb{C}$ show that$\\\begin{align*}|v| + |v + w| + |w + z| + |2 + z| \geqslant 2.\end{align*}

Re: HSC 2016 4U Marathon $\noindent Writing the two complex numbers in modulus--argument form one has: $w = |w| \, \mbox{cis} \, \phi$ and $z = |z| \, \mbox{cis} \, \theta$. Now let $w$ form one side of the triangle and $z$ the other such that the included angle between these two sides is...

Re: HSC 2016 4U Marathon It's working again (well, at least for me it's working again)

Re: MX2 2016 Integration Marathon $\noindent We will use IBP. We begin by noting that \begin{align*}\frac{d}{dx} \left [\ln \left (\sqrt{1 + x} + \sqrt{1 - x} \right ) \right ] &= \frac{2 \sqrt{1 - x} - 2 \sqrt{1 + x}}{4 \sqrt{1 - x} \sqrt{1 + x}} \cdot \frac{1}{\sqrt{1 + x} + \sqrt{1 -...

Re: MX2 2016 Integration Marathon Correct!

Re: MX2 2016 Integration Marathon $\noindent And now for some integration `humour.' $\noindent Find $\int \frac{dx}{dx}.

Re: MX2 2016 Integration Marathon $\noindent \textbf{Next Question} $\noindent Find $\int \frac{\ln(1 + \sin^2 x)}{\sin^2 x} dx.