Search results

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!

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: 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...