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Re: MX2 2016 Integration Marathon $IBP gives I_n = x \arcsin^n x - n \int \frac{x}{\sqrt{1 - x^2}} \arcsin^{n - 1} x \, dx $By noting $\int \frac{x}{\sqrt{1 - x^2}} dx = -\sqrt{1 - x^2} + \cal{C}$ (use the substitution $x = \sin \theta$ to show this)$ $IBP again yields$ I_n = x...

Re: Announcement from BOSTES - significant change to calculus courses Please do when you get the time. I am interested to hear what the current thinking on these changes from other teachers is.

Re: MX2 2015 Integration Marathon $Let $I = \int^{\frac{5\pi}{2}}_{\frac{\pi}{2}} \frac{{\rm{e}}^{\tan^{-1}(\sin x)}}{{\rm{e}}^{\tan^{-1}(\sin x)} + {\rm{e}}^{\tan^{-1}(\cos x)}} \, dx. $Now $\\\begin{align*}I &= \int^{0}_{\frac{\pi}{2}} \frac{{\rm{e}}^{\tan^{-1}(\sin...

Re: MX2 2015 Integration Marathon \int^\pi_0 \frac{d\theta}{(2 - \cos \theta)^2}

Re: MX2 2015 Integration Marathon $Let $ I_n = \int^1_0 {\rm{e}}^x ( x - 1)^n \, dx, \quad n \geqslant 0.\\$On integrating by parts, we have$ \begin{align*}I_n &= \left [{\rm{e}}^x (x - 1)^n \right ]^1_0 - \int^1_0 {\rm {e}}^x n (x - 1)^{n - 1} \, dx\\&= -(-1)^n - n \int^1_0 {\rm {e}}^x...

Re: MX2 2015 Integration Marathon $(a) $ \int \left (\frac{\tan^{-1} x}{x - \tan^{-1} x} \right )^2 \,dx \qquad $(b) $ \int^1_{-1} \tan^{-1} \left ({\rm{e}}^{\sin x} \right ) \, dx.

Re: MX2 2015 Integration Marathon $In addition to juantheron's solution, I will show how to evaluate this integral using three other methods. Let$\\I = \int \frac{x^2}{(x \sin x + \cos x)^2} \, dx. $\textbf{Method I}$ $Observe that $\frac{d}{dx} (x \sin x + \cos x) = x \cos x.$ So...

Re: MX2 2015 Integration Marathon Many moons ago - 1990

Re: MX2 2015 Integration Marathon $For convenience let us shift the index $k$ by one, so$\\f(k - 1) = \int^1_{-1} \frac{\sqrt{1 - x^2}}{\sqrt{k} - x} \, dx.$ Here $k \in \mathbb{N}. $Now consider the indefinite integral $I = \int \frac{\sqrt{1 - x^2}}{\sqrt{k} - x} \, dx...

Re: MX2 2015 Integration Marathon $Another way to get there is as follows. Let $I = \int \frac{\sin^{-1} x}{x^2} dx$ and let $y = \sin^{-1} x \,\, \Rightarrow \,\, x = \sin y.$ So $dx = \cos y \, dy$ and we have$ \begin{align*}I &= \int \frac{y}{\sin^2 y} \cos y \, dy\\&= \int y \cot y \csc y...

Agree that there are probably not enough qualified teachers readily avaiable to teach a real physics course in many of our schools but we at least should try. And as for wanting more students/girls to take the course, that is political. Not many students these days take Latin, but no one seems...

I would like to know what others think of the proposal currently being put forward by the Board for possible changes in the HSC physics course (see: http://www.boardofstudies.nsw.edu.au/australian-curriculum/pdf_doc/physics-st6-draft-writing-brief-2015.pdf). They are currently proposing 3...

Re: MX2 2015 Integration Marathon $(a)$ \int \frac{\sin^{-1} x}{x^2} dx \qquad $(b)$ \int \left (\sin^{-1} x \right )^3 dx.

Re: MX2 2015 Integration Marathon $Using$\\\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B},$ after rearranging we have$\\\tan A + \tan B = \frac{\tan A - \tan B}{\tan (A - B)} - 1.\\$Setting $ A = x - a$ and $B = x - b$ we have$\\\begin{align*}\tan (x - a) \tan (x - b) &= \frac{\tan(x -...

Re: MX2 2015 Integration Marathon $Now for the function $f$, as $f'(x) = 3x^2 + 3 > 0$ for all $x$ it is an increasing function on its domain and is therefore one-to-one and hence possesses an inverse. We begin by noting that$\\f(-1) = 0 \Rightarrow f^{-1}(0) = -1 \quad \mbox{and} \quad f(0) = 4...

Re: MX2 2015 Integration Marathon $A small \textbf{Lemma} to begin with\\If $f^{-1}(x)$ is invertible on the interval $a \leqslant x \leqslant b$, then\\$\int^b_a f^{-1}(x) \, dx = b \cdot f^{-1}(b) - a \cdot f^{-1}(a) - \int^{f^{-1}(b)}_{f^{-1}(a)} f(x) \, dx.\\$\textbf{Proof}$\\$Let $y =...

The dropping of first-order differential equations for conics perhaps :lol:

Perhaps, but why the rush to try and cover material at the secondary level which many will simply see again at university? And from some of the new topics being proposed one will only be able to treat them at a very superficial level anyway (difference and differential equations for example)...

Sad but true!

The percentage of people who went on and finished the last two years of secondary school was for much to the 20th century very low. Those who did go on and complete the last two years of school were those who were usually the most capable and were motivated to do so. Nowadays, all students are...