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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}.

Re: MX2 2016 Integration Marathon $\noindent Provided $n \neq \frac{1}{2}$. When $n = \frac{1}{2}$ one has:$\\-\frac{1}{2} \ln |1 - x^2| + \cal{C}.

Re: MX2 2016 Integration Marathon $\noindent \textbf{Next Question}$ $\noindent Briefly explain what is wrong with the following ``proof''. $\noindent If $f(x) \neq 0$, from integration by parts we have$\\\begin{align*}\int \frac{f'(x)}{f(x)} \, dx = \frac{f(x)}{f(x)} + \int \frac{f(x)...

Re: MX2 2016 Integration Marathon $\noindent Let $x = \tan^6 \theta, dx = 6 \tan^5 \theta \sec^2 \theta \, d\theta$. So\\\begin{align*}\int \frac{dx}{\sqrt{x}(1 + \sqrt[3]{x})} &= \int \frac{6 \tan^5 \theta \sec^2 \theta}{\tan^3 \theta (1 + \tan^2 \theta)} \, d\theta\\ &= 6 \int \tan^2 \theta...

Re: Extracurricular Integration Marathon $\noindent Yes, $\frac{5\pi^2}{96}$ is the correct answer but surely you didn't just guess it?

Re: MX2 2016 Integration Marathon $\noindent Alternatively a substitution of $x = \sin \theta$ followed by recalling that $\\\begin{align*}\sqrt{\frac{1 - \sin \theta}{1 + \sin \theta}} &= \tan \left (\frac{\theta}{2} + \frac{\pi}{4} \right )\end{align*}\\$ would work as well, though perhaps...