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Re: MX2 2016 Integration Marathon $\noindent The integral does not converge for all values of $ a.$ For convergence I will assume $-\frac{\pi}{2} < a < \frac{\pi}{2}. $\noindent Let $I = \int^a_0 \frac{x}{\cos x \cos (a - x)} \, dx.$ Using the well-known result of $ \int^a_0 f(x) \, dx...

Re: MX2 2016 Integration Marathon $\noindent Alternatively, let $x = u^2, dx = 2u \, du$ so that$ \begin{align*}\int \frac{9 + 6\sqrt{x} + x}{4\sqrt{x} + x} \, dx &= 2 \int \frac{9 + 6u + u^2}{4 + u} \, du\\&= 2 \int \frac{(u + 4)(u + 2) + 1}{u + 4} \, du\\&= 2 \int \left (u + 2 +...

Re: MX2 2016 Integration Marathon $\noindent In a similar vein to my previous question, find $\int \left (\frac{1 - \tan x}{1 - \sec x} \right ) \mbox{e}^x \, dx.

Re: MX2 2016 Integration Marathon Yes, I agree, your method is definitely more elegant.

Re: MX2 2016 Integration Marathon I see InteGrand beat me to it.....

Re: MX2 2016 Integration Marathon $\noindent For those after a more systematic approach, this is what I used. Observe that $\\\begin{align*}\frac{d}{dx} \left (f(x) e^x \right ) = \left (f(x) + f'(x) \right ) e^x\\\Rightarrow \int \left (f(x) + f'(x) \right ) e^x \, dx = f(x) e^x +...

Re: MX2 2016 Integration Marathon $\noindent \textbf{Next Question}$ $\noindent Find $ \int \left (\frac{1 - \sin x}{1 - \cos x} \right ) e^x \, dx

Re: MX2 2016 Integration Marathon $\noindent Yes but each of your approaches were slightly different and leehuan accidently dropped a factor of 12 in the first term of the final answer.

Re: MX2 2016 Integration Marathon $\noindent \textbf{Next Question}$ $\noindent Find $\int \frac{\sqrt[4]{1 + \sqrt[3]{x}}}{x} \, dx

Re: MX2 2016 Integration Marathon $\noindent \textbf{Next Question} -- something a little easier $\noindent Find $ \int \cos^{12}(5x) \sin^3 (5x) \, dx.

Re: MX2 2016 Integration Marathon $\noindent There is probably a more elegant way to do this, and it is a bit of a slog fest, but this is what I did. I started by rewriting the integrand as follows:$ \begin{align*}\int \frac{x^2 - 7x + 3}{\sqrt{x^2 + 2x + 7}} \, dx&=\int \frac{(x + 1)^2 -...

Re: MX2 2016 Integration Marathon $\noindent Alternatively, if one recognises$\\\begin{align*}\frac{1}{x - x^{2016}} = \frac{1}{x(1 - x^{2015})} = \frac{1}{x} + \frac{x^{2014}}{1 - x^{2015}},\end{align*} $\noindent One has\\\begin{align*}\int \frac{dx}{x - x^{2016}} &= \int \frac{dx}{x} +...

Re: MX2 2016 Integration Marathon $\noindent Correct. For the benefit of others, from Paradoxia's suggestion we have \begin{align*}\int \frac{dx}{x - x^{2016}} &= \int \frac{dx}{x^{2016}(x^{-2015} - 1)}\\&= \int \frac{x^{-2016}}{x^{-2015} - 1} dx.\end{align*} $\noindent Let $u =...

Re: MX2 2016 Integration Marathon $\noindent Next question for the 2016ers. Find $\int \frac{dx}{x - x^{2016}}.

Re: MX2 2016 Integration Marathon $\noindent Use IBP twice. Doing so we have$\\\begin{align*}\int \sin (\ln x) \, dx &= x \sin (\ln x) - \int \cos (\ln x) \, dx\\&=x \sin (\ln x) - x \cos (ln x) - \int \sin (\ln x) \, dx\\\Rightarrow 2 \int \sin (\ln x) \, dx &= x \sin (\ln x) - x \cos (\ln...

Re: MX2 2016 Integration Marathon $\noindent For all integers, in which case $m = n = 0$ will also need to be considered separately, or for just positive integers $ m$ and $n$ only?

Re: MX2 2016 Integration Marathon $\noindent Or for a third approach which also avoids the dreaded partial fractions, start with the standard substitution of $u^2 = \tan x$ to arrive at$ \int \sqrt{\tan x} \, dx = \int \frac{2u^2}{1+u^4} du. $\noindent Now rather than launching head...

Re: MX2 2016 Integration Marathon $No need to scare the children away so soon!$ $\noindent We begin by noting that$\\\begin{align*}\sqrt{\tan x} &= \frac{1}{2} \sqrt{\tan x} + \frac{1}{2} \sqrt{\tan x} + \frac{1}{2} \sqrt{\cot x} - \frac{1}{2} \sqrt{\cot x}\\&= \frac{1}{2} \left (\sqrt{\tan...

Re: HSC 2016 4U Marathon $Not as slick as Paradoxica's solution, here I give another approach. $\noindent By inspection $-1$ is a root of the given polynomial (which is symmetric). Thus $ \begin{align*}ax^3 + bx^2 + bx + a = (x + 1)(a x^2 + (b - a) x + a).\end{align*} $For the remaining two...

Re: MX2 2016 Integration Marathon Oh yes. Silly me!