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Because as a name we all know 4 Unit is so such better than Extension 2 :)

Re: MX2 2016 Integration Marathon $\noindent \textbf{Next Question}$ $\noindent Find $\int \frac{dx}{(x + 2)^{\frac{6}{5}} (x - 7)^{\frac{4}{5}}}.

Re: MX2 2016 Integration Marathon $\noindent Need to keep the train moving right along with a \textbf{New Question}$ $\noindent Evaluate $\int^1_0 \frac{x}{(x^2 + 3) \sqrt{x^2 + 2}} \, dx.

Re: MX2 2016 Integration Marathon $\noindent Ah yes, the second one just equals $\sin^{-1} x$. So for completeness$\\\begin{align*}\int \frac{\sqrt{1-x^2}}{x^2 + 1} \, dx = \sqrt{2} \tan^{-1} \left ( \frac{\sqrt{2} x}{\sqrt{1 - x^2}} \right ) + \sin^{-1} x + \cal{C}\end{align*}

Re: MX2 2016 Integration Marathon $\noindent Let $x = \sin \theta, dx = \cos \theta \, d\theta.$ So$\\\begin{align*}\int \frac{\sqrt{1-x^2}}{x^2+1} \, dx &= \int \frac{\cos^2 \theta}{\sin^2 \theta + 1} \, d\theta\\&=\int \frac{\cos^2 \theta}{\sin^2 \theta + (\sin^2 \theta + \cos^2 \theta)}...

Re: MX2 2016 Integration Marathon $\noindent \textbf{Next Question}$ $\noindent Find $\int \sqrt{x^2 + x^{-2} - 1} \, dx.

Re: MX2 2016 Integration Marathon $\noindent Paradoxica, I think this question should really be in the Extracurricular Integration Marathon, but here goes anyway. $\noindent Decomposing the rational part of the integrand into partial fractions, while in principal is not hard, in this...

Re: Extracurricular Integration Marathon $\noindent \textbf{Two New Questions}$ $\noindent Evalaute$\\$(a) $\begin{align*}\int^{\frac{\pi}{2}}_0 \cot x \ln (\sec x) \, dx \,\,\,\,\,\,\,&\mbox{(b)} \,\, \int^1_0 \frac{\tan^{-1} (\sqrt{2 + x^2})}{(1 + x^2) \sqrt{2 + x^2}} \, dx\end{align*}

Re: Extracurricular Integration Marathon \begin{align*}\lim _{n\to \infty} \int _0 ^\infty \frac{\mathrm{e}^{-t}\cos t}{\frac{1}{n} + nt^2} \text{ d}t &= \lim_{n \rightarrow \infty} \frac{1}{n} \int^\infty_0 \frac{e^{-t} \cos t}{\frac{1}{n^2} + t^2} \, dt\\&= \lim_{n \rightarrow \infty}...

Re: Extracurricular Integration Marathon $\noindent Let $x = e^u, dx = e^u \, du$. For the limits of integration we have: $x = 0, u \rightarrow -\infty$ and $x \rightarrow \infty, u \rightarrow \infty.$ So \begin{align*} \int^\infty_0 f \left (x + \frac{1}{x} \right ) \frac{\ln x}{x} \, dx &=...

Re: Extracurricular Integration Marathon $\noindent We start first with a preliminary result. For $n \geqslant 1$ using IBP it can be shown that$\\\begin{align*}\int^1_0 x^n \ln x \, dx &= - \frac{1}{(n + 1)^2}.\end{align*} $\noindent Now, let $u = 1 - e^{-x^2}, du = 2x e^{-x^2} \, dx...

If you are interested in studying real physics and chemistry I would recommend doing IB over HSC. What is currently presented as physics and chemistry at HSC-level is not physics or chemistry at all but some sort of watered down drivel almost devoid of any real science. Of course, by choosing to...

Re: MX2 2016 Integration Marathon $\noindent \textbf{Next Question} $\noindent Evaluate $\int^{\frac{\pi}{4}}_{-\frac{\pi}{4}} \frac{x^4}{1 + \tan x + \sec x} \, dx.

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

Re: MX2 2016 Integration Marathon $\noindent A final answer of $-\frac{4}{5 \sqrt{5}} \ln \left |\frac{\sqrt{5} + \cos x - 2 \sin x}{\sin x + 2 \cos x} \right | - \frac{1}{5} \cos x - \frac{2}{5} \sin x + \cal{C}$ looks pretty ``clean'' to me ;-)$

Re: MX2 2016 Integration Marathon $\noindent And is exactly why I selected the limits of integration I did!$

And <a href="https://www.researchgate.net/publication/243715872_Flexible_Faraday_Cage_with_a_Twist_Surface_Charge_on_a_Mobius_Strip">here</a> is a very simple physical demonstration of the one-sided property of the Möbius strip that makes use of surface charge and can easily be done in a high...

Re: MX2 2016 Integration Marathon \begin{align*} \int \frac{dx}{\sin (x - \frac{\pi}{3} ) \cdot \cos x} &= \int \frac{2 \sec x}{\sin x - \sqrt{3} \cos x} \, dx = 2 \int \frac{\sec^2 x}{\tan x - \sqrt{3}} \, dx\end{align*} $\noindent Now let $u = \tan x, du = \sec^2 \, dx$...

Re: MX2 2016 Integration Marathon $\noindent Particular care needs to be taken with the limits of integration. Would you care to post a solution for the benefit of others?$

Re: MX2 2016 Integration Marathon $\noindent I think we need a \textbf{New Question}$ $\noindent Evaluate $\int^{\sqrt{3}}_0 \sin^{-1} \left (\frac{2x}{1 + x^2} \right ) \, dx