Search results

One of the three options being proposed for physics (and chemistry as well) late last year which BOSTES were seeking feedback on, was a return to a standard, introductory-level physics course (essentially something akin to what we had pre-2001). Although the proposal was a little light on...

I am guessing many students find first-year physics harder than 4 Unit Maths since it is most likely their first ever exposure to real physics. There is absolutely no relation between the real physics taught at universities with that light weigh puff piece known as HSC Physics. The latter is...

Re: Help i'm trapped in the title and can only speak once every 2 years $\noindent I will assume $n \neq 0. \begin{align*}I &= \int 20n x^{5n-1} \tan^{-1}\left( \frac{(x^n-1)x^n(x^n+1)}{(x^{2n}-\sqrt{3}x^n+1)(x^{2n}+\sqrt{3}x^n+1)} \right)\,dx\\&= \int 20 n x^{5n - 1} \tan^{-1} \left...

Re: MX2 2016 Integration Marathon $\noindent For those who wish to keep it purely real, various product-to-sum and sum-to-product identities for the trig functions can be used to find each of the sums. For the general case we will consider $\frac{\sum^{n - 1}_{k = 1} \sin k\theta}{\sum^{n -...

Re: MX2 2016 Integration Marathon $\noindent \textbf{Next Question} $\noindent Find $\int \frac{\sin x + \sin 2x + \sin 3x + \sin 4x}{\cos x + \cos 2x + \cos 3x + \cos 4x} \, dx

Re: MX2 2016 Integration Marathon $\noindent I will show the second result directly. Let$ I_k = \int^1_0 x^k (1 - x)^{n - k} \, dx. $\noindent Now $I_0 = \int^1_0 (1 - x)^n \, dx = \int^1_0 x^n \, dx = \frac{1}{n + 1}. \begin{align*}I_k &= \left [\frac{x^{k + 1}}{k + 1} (1 - x)^{n -...

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

Re: Help i'm trapped in the title and can only speak once every 2 years $\noindent Please let me know if there is a simpler way to do this one. $\noindent Let $V = \int \tan^{-1} (2^{2n - 1} x^2) \, dx.$ Now let $2^{2n - 1} x^2 = \tan \theta, 2^{2n} x \, dx = \sec^2 \theta \, d\theta$...

Re: MX2 2016 Integration Marathon $\noindent Of course this one can be directly found using a partial fraction decomposition, but to do so is a pretty hard slog. Instead, we will first try and manipulate the integrand a bit. \begin{align*}\int \frac{dx}{(x^4 - 1)^2} &= \frac{1}{4} \int...

Re: MX2 2016 Integration Marathon $\noindent \textbf{Next Question}$ $\noindent Find $ \int \frac{\sin x}{\sin (x - a) \sin (x - b)} \, dx, \quad a \neq b.

Re: MX2 2016 Integration Marathon $\noindent Consider the finite product term first. Note that$ \begin{align*} \sin x &= 2 \sin \frac{x}{2} \cos \frac{x}{2} = 2^1 \sin \frac{x}{2^1} \cos \frac{x}{2^1}\\&= 2 \cdot 2 \sin \frac{x}{4} \cos \frac{x}{4} \cdot \cos \frac{x}{2} = 2^2 \sin...

Re: MX2 2016 Integration Marathon $\noindent \textbf{Next Question}$ $\noindent Evaluate $\int^5_0 \frac{|x - 1|}{|x - 2| + |x - 4|} \, dx.

Re: MX2 2016 Integration Marathon $\noindent In case anyone would like to see how this one is done. $\noindent Let $I = \int_0^1{\frac{1}{(x^2-x+1)(e^{2x-1}+1)}} \, dx $\noindent Using a so-called ``boarder flip'' we have:$ \begin{align*}I &= \int^1_0 \frac{dx}{[(1 - x)^2 - (1 - x)...

Re: MX2 2016 Integration Marathon \begin{align*}\int \frac{2x - 3}{x(x - 1)(x- 2)(x - 3) + 2} \, dx &= \int \frac{2x - 3}{x (x - 3) \cdot (x - 2)(x - 3) + 2} \, dx\\&= \int \frac{2x - 3}{(x^2 - 3x)(x^2 - 3x + 2) + 2} \, dx\end{align*} $\noindent Let $u = x^2 - 3x, du = (2x - 3) \, dx$ so...

Re: MX2 2016 Integration Marathon $\noindent \textbf{Next Question}$ $\noindent Find $\int \frac{\cos 5x - \cos 4x}{1 - \cos 3x} \, dx. $\noindent I hope this one requires a little bit more work than ``inspection'' to find.$

Re: MX2 2016 Integration Marathon $\noindent If possible, could someone please pick up and post a copy of the questions to this forum after it has been held. I would be interested in seeing the types of questions they ask. Thanks.$

Re: MX2 2016 Integration Marathon $\noindent \textbf{Next Question} $\noindent Find $\int \frac{\cos 5x + \cos 4x}{1 - 2 \cos 3x} \, dx$.

Re: MX2 2016 Integration Marathon $\noindent Let $n = -k$ where $k \in \mathbb{N}. \begin{align*}\int \frac{dx}{1 - \sqrt[n]{x}} &= \int \frac{dx}{1 - x^{-1/k}} = \int \frac{x^{1/k}}{x^{1/k} - 1} \, dx.\end{align*} $\noindent Now let $u = x^{1/k}, dx = k u^{k - 1} \, du.$ Thus$...

Re: MX2 2016 Integration Marathon Oh yes. Silly me

Re: MX2 2016 Integration Marathon $\noindent Provided $n \neq 0$. When $n = 0$ we just have $\frac{1}{\sqrt{2}} \ln |x| + \cal{C}.