HSC 2017 MX2 Integration Marathon (archive) (3 Viewers)

leehuan · Nov 3, 2016

Re: HSC 4U Integration Marathon 2017

pikachu975 said:
How do you already know 4u integration wtf

It's just integration. It's the easier one to self teach.

His skill is in that he tackled a decently harder one.

Drsoccerball · Nov 3, 2016

Re: HSC 4U Integration Marathon 2017

jathu123 said:
started ages ago, from watching vids and looking at other peoples working out on marathons. (thats partially how i grasped integration by substitution lol, by looking at people making dx the subject etc). Then participating in marathons strengthens the understanding as well as helping learn new stuff.

Dividing by tan^2 is something you pretty much only see in the marathons. Can see you've been working hard keep going.

Paradoxica · Nov 3, 2016

Re: HSC 4U Integration Marathon 2017

jathu123 said:
lmao no way, I shouldn't be even compared to him

Become my student and one day you will best me.

Drsoccerball · Nov 4, 2016

Re: HSC 4U Integration Marathon 2017

Paradoxica said:
Become my student and one day you will best me.

1 Paradoxica is enough tyvm.

juantheron · Nov 22, 2016

Re: HSC 4U Integration Marathon 2017

$Evaluation of $\int\frac{3+4\cos x}{(4+3\cos x)^2}dx$

$Evaluation of $\int^{\frac{\pi}{4}}_{0}(\cos 2x)^{\frac{3}{2}}\cdot \cos xdx$

Paradoxica · Dec 6, 2016

Re: HSC 4U Integration Marathon 2017

$\int_{-1}^1 \sin{(\pi |x|)} \sin^{-1}{(\sqrt{|x|})} \text{d}x$

Lugia101101 · Dec 6, 2016

Re: HSC 4U Integration Marathon 2017

I didn't actually do 4U, so there's probably a method to do it properly, but:
$\begin{align*}\text{So }&\int_{-1}^{1}{\sin{(\pi |x|)}\sin^{-1}{\sqrt{|x|}}\text{d}x}\\ =&2\int_{0}^{1}{\sin{(\pi x)}\sin^{-1}{\sqrt{x}}\text{d}x}\\ =&\frac{2}{\pi}\big[\cos{(\pi x)}\sin^{-1}{x}\big]_{1}^{0}+\frac{1}{\pi}\int_{0}^{1}{\frac{\cos{(\pi x)}}{\sqrt{x-x^2}}\text{d}x}\\ =&1+\frac{1}{\pi}\int_{0}^{1}{\frac{\cos{(\pi x)}}{\sqrt{x-x^2}}\text{d}x}\end{align*}$
$\begin{align*}\text{Now as }\cos{(\pi x)}=&-\cos{(\pi(1-x))}\\ \text{is }&\text{positive for }0<x<\frac{1}{2}\\ &\text{negative for }\frac{1}{2}<x<1\end{align*}$
$\begin{align*}\text{And }\sqrt{x-x^2}&=\sqrt{(1-x)-(1-x)^2}\\ &\text{is positive for }0<x<1\end{align*}$
$\int_{0}^{\frac{1}{2}}{\cos{(\pi x)}\text{d}x}=-\int_{\frac{1}{2}}^{1}{\cos{(\pi x)}\text{d}x}$
$\text{So }1+\frac{1}{\pi}\int_{0}^{1}{\frac{\cos{(\pi x)}}{\sqrt{x-x^2}}\text{d}x}=1+\frac{0}{\pi}=1$
$\therefore\int_{-1}^{1}{\sin{(\pi |x|)}\sin^{-1}{\sqrt{|x|}}\text{d}x}=1$

leehuan · Dec 7, 2016

Re: HSC 4U Integration Marathon 2017

juantheron said:
$Evaluation of $\int^{\frac{\pi}{4}}_{0}(\cos 2x)^{\frac{3}{2}}\cdot \cos xdx$

$\sqrt{2}\sin x = \sin \theta \implies \cos x \, dx =\frac1{\sqrt2}\cos \theta \, d\theta$

$\begin{align*}&\quad \int_0^{\frac{\pi}4}(\cos 2x)^{\frac32}\cos x\, dx\\ &= \int_0^{\frac{\pi}4}(1-2\sin^2x)^{\frac32}\cos x\, dx\\ &= \frac{1}{\sqrt2}\int_0^{\frac{\pi}2}(1-\sin^2\theta)^{\frac{3}{2}} \cos \theta \, d \theta\\ &= \frac{1}{\sqrt2}\int_0^{\frac{\pi}2}\cos^4\theta\,d\theta \end{align*}\\ \text{which can be tackled by a reduction formula.}$

Paradoxica · Dec 8, 2016

Re: HSC 4U Integration Marathon 2017

Let's try that again.

$\int_{-1}^1 \sin{(\pi |x|)} \sin^{-1}{(\sqrt{|x|})} \text{d}x$

$$\noindent You must use the fact that $\sin^{-1}{\sqrt{1-u^2}} = \cos^{-1}{u}$ for all $0\le u\le 1$

Lugia101101 · Dec 8, 2016

Re: HSC 4U Integration Marathon 2017

$\begin{align*}\text{So }&\int_{-1}^{1}{\sin{(\pi|x|)}\sin^{-1}{\sqrt{|x|}}\text{d}x}\\ =&2\int_{0}^{1}{\sin{(\pi x)}\sin^{-1}{\sqrt{x}}\text{d}x}\\ \text{Let }&\pi x=\cos^{-1}{u}\\ &\text{d}x=\frac{-\text{d}u}{\pi\sqrt{1-u^2}}\\ &x=0\implies u=1\\ &x=1\implies u=-1\\ =&2\int_{1}^{-1}{\sin{(\cos^{-1}{u})}\sin^{-1}{\sqrt{\frac{1}{\pi}\cos^{-1}{u}}}\frac{-\text{d}u}{\pi\sqrt{1-u^2}}}\end{align*}$
$\begin{align*}\text{Now }\sin^{-1}{\sqrt{1-u^2}}&=\cos^{-1}{u}\\ \sqrt{1-u^2}&=\sin{(\cos^{-1}{u})}\end{align*}$
$\begin{align*}\text{So }&2\int_{1}^{-1}{\sin{(\cos^{-1}{u})}\sin^{-1}{\sqrt{\frac{1}{\pi}\cos^{-1}{u}}}\frac{-\text{d}u}{\pi\sqrt{1-u^2}}}\\ =&2\int_{1}^{-1}{\sqrt{1-u^2}\sin^{-1}{\sqrt{\frac{1}{\pi}\cos^{-1}{u}}}\frac{-\text{d}u}{\pi\sqrt{1-u^2}}}\\ =&\frac{2}{\pi}\int_{-1}^{1}{\sin^{-1}{\sqrt{\frac{1}{\pi}\cos^{-1}{u}}}\text{d}u}\end{align*}$
$\begin{align*}\text{Let }y&=\sin^{-1}{\sqrt{\frac{1}{\pi}\cos^{-1}{u}}}\\ \sin^{2}{y}&=\frac{\cos^{-1}{u}}{\pi}\\ u&=\cos{(\pi\sin^{2}{y})}\end{align*}$
$\text{So }\frac{2}{\pi}\int_{-1}^{1}{\sin^{-1}{\sqrt{\frac{1}{\pi}\cos^{-1}{u}}}\text{d}u}\equiv\frac{2}{\pi} \left(\int_{0}^{\frac{\pi}{4}}{\cos{(\pi\sin^{2}{y})} \text{d}y}+\frac{\pi}{2}+\int_{\frac{\pi}{4}}^{ \frac{\pi}{2}}{\cos{(\pi\sin^{2}{y})} \text{d}y}\right)*$
$*\text{Using equivalent areas of inverse functions}$
$\begin{align*}\text{And as }&-\cos{(\pi\sin^{2}{(\frac{\pi}{2}-y)})}\\ =&-\cos{(\pi\cos^{2}{y})}\\ =&-\cos{(\pi-\pi\sin^{2}{y})}\\ =&\cos{(\pi\sin^{2}{y})}\end{align*}$
$\int_{0}^{\frac{\pi}{4}}{\cos{(\pi\sin^{2}{y})} \text{d}y}=-\int_{\frac{\pi}{4}}^{ \frac{\pi}{2}}{\cos{(\pi\sin^{2}{y})} \text{d}y}$
$\begin{align*}\text{So }\frac{2}{\pi} \left(\int_{0}^{\frac{\pi}{4}}{\cos{(\pi\sin^{2}{y})} \text{d}y}+\frac{\pi}{2}+\int_{\frac{\pi}{4}}^{ \frac{\pi}{2}}{\cos{(\pi\sin^{2}{y})} \text{d}y}\right)&=\frac{2}{\pi}\left(\frac{\pi}{2}\right)\\ &=1\end{align*}$
$\therefore\int_{-1}^{1}{\sin{(\pi|x|)}\sin^{-1}{\sqrt{|x|}}\text{d}x}=1$

Paradoxica · Dec 8, 2016

Re: HSC 4U Integration Marathon 2017

Lugia101101 said:
$\begin{align*}\text{So }&\int_{-1}^{1}{\sin{(\pi|x|)}\sin^{-1}{\sqrt{|x|}}\text{d}x}\\ =&2\int_{0}^{1}{\sin{(\pi x)}\sin^{-1}{\sqrt{x}}\text{d}x}\\ \text{Let }&\pi x=\cos^{-1}{u}\\ &\text{d}x=\frac{-\text{d}u}{\pi\sqrt{1-u^2}}\\ &x=0\implies u=1\\ &x=1\implies u=-1\\ =&2\int_{1}^{-1}{\sin{(\cos^{-1}{u})}\sin^{-1}{\sqrt{\frac{1}{\pi}\cos^{-1}{u}}}\frac{-\text{d}u}{\pi\sqrt{1-u^2}}}\end{align*}$
$\begin{align*}\text{Now }\sin^{-1}{\sqrt{1-u^2}}&=\cos^{-1}{u}\\ \sqrt{1-u^2}&=\sin{(\cos^{-1}{u})}\end{align*}$
$\begin{align*}\text{So }&2\int_{1}^{-1}{\sin{(\cos^{-1}{u})}\sin^{-1}{\sqrt{\frac{1}{\pi}\cos^{-1}{u}}}\frac{-\text{d}u}{\pi\sqrt{1-u^2}}}\\ =&2\int_{1}^{-1}{\sqrt{1-u^2}\sin^{-1}{\sqrt{\frac{1}{\pi}\cos^{-1}{u}}}\frac{-\text{d}u}{\pi\sqrt{1-u^2}}}\\ =&\frac{2}{\pi}\int_{-1}^{1}{\sin^{-1}{\sqrt{\frac{1}{\pi}\cos^{-1}{u}}}\text{d}u}\end{align*}$
$\begin{align*}\text{Let }y&=\sin^{-1}{\sqrt{\frac{1}{\pi}\cos^{-1}{u}}}\\ \sin^{2}{y}&=\frac{\cos^{-1}{u}}{\pi}\\ u&=\cos{(\pi\sin^{2}{y})}\end{align*}$
$\text{So }\frac{2}{\pi}\int_{-1}^{1}{\sin^{-1}{\sqrt{\frac{1}{\pi}\cos^{-1}{u}}}\text{d}u}\equiv\frac{2}{\pi} \left(\int_{0}^{\frac{\pi}{4}}{\cos{(\pi\sin^{2}{y})} \text{d}y}+\frac{\pi}{2}+\int_{\frac{\pi}{4}}^{ \frac{\pi}{2}}{\cos{(\pi\sin^{2}{y})} \text{d}y}\right)*$
$*\text{Using equivalent areas of inverse functions}$
$\begin{align*}\text{And as }&-\cos{(\pi\sin^{2}{(\frac{\pi}{2}-y)})}\\ =&-\cos{(\pi\cos^{2}{y})}\\ =&-\cos{(\pi-\pi\sin^{2}{y})}\\ =&\cos{(\pi\sin^{2}{y})}\end{align*}$
$\int_{0}^{\frac{\pi}{4}}{\cos{(\pi\sin^{2}{y})} \text{d}y}=-\int_{\frac{\pi}{4}}^{ \frac{\pi}{2}}{\cos{(\pi\sin^{2}{y})} \text{d}y}$
$\begin{align*}\text{So }\frac{2}{\pi} \left(\int_{0}^{\frac{\pi}{4}}{\cos{(\pi\sin^{2}{y})} \text{d}y}+\frac{\pi}{2}+\int_{\frac{\pi}{4}}^{ \frac{\pi}{2}}{\cos{(\pi\sin^{2}{y})} \text{d}y}\right)&=\frac{2}{\pi}\left(\frac{\pi}{2}\right)\\ &=1\end{align*}$
$\therefore\int_{-1}^{1}{\sin{(\pi|x|)}\sin^{-1}{\sqrt{|x|}}\text{d}x}=1$

This is not using the fact.

And most of that working vanishes with just a single observation.

leehuan · Dec 8, 2016

Re: HSC 4U Integration Marathon 2017

$\text{Let }I=\int_{-1}^1\sin(\pi |x|)\arcsin\sqrt{|x|}\,dx\text{ then we have}$

$\begin{align*}I&=2\int_0^1\sin (\pi x)\arcsin\sqrt{x}\,dx\\ &= 2\int_0^1\sin (\pi x)\arcsin\sqrt{1-\left(\sqrt{1-x}\right)^2}\,dx\\ &= 2\int_0^1 \sin (\pi x)\arccos \sqrt{1-x}\, dx\end{align*}\\ \text{noting that }0\le x \le 1\implies 0\le 1-x \le 1 \implies \sqrt{1-x}\le 1$

$\text{By considering the substitution }u=1-x\\ I=2\int_0^1\sin (\pi u)\arccos\sqrt{u}\,du\\ \text{noting }\sin(\pi - \alpha)=\sin \alpha$

$\text{Hence upon addition and noting dummy variables in integration}\\ \begin{align*}2I&=2\int_0^1(\sin (\pi x))\left(\arcsin \sqrt{x}+\arccos \sqrt {x}\right)dx\\ I&=\frac{\pi}{2}\int_0^1\sin(\pi x)\,dx\\&= 1 \end{align*}$

jathu123 · Dec 12, 2016

Re: HSC 4U Integration Marathon 2017

juantheron said:
$Evaluation of $\int\frac{3+4\cos x}{(4+3\cos x)^2}dx$

$I=\int \frac{(3+4\cos x)}{(4+3\cos x)^2}$ d$x\\ $ Let $t =\tan \left ( \frac{x}{2} \right )\\ $ d$t=\frac{1}{2}\sec^2\left ( \frac{x}{2} \right )$ d$x\\\Rightarrow $ d$x=\frac{2}{1+t^2}$ d$t \\I= 2\int \frac{\left (\frac{3+3t^2}{1+t^2}+\frac{4-4t^2}{1+t^2} \right )}{\left ( \frac{4+4t^2}{1+t^2} +\frac{3t-t^2}{1+t^2}\right )^2}\cdot \frac{2}{1+t^2}$ d$t \\ =2\int \frac{7-t^2}{(7+t^2)^2}$ d$t\\= 2\int \frac{1}{7+t^2}$ d$t-2\int \frac{2t^2}{(7+t^2)^2}$ d$t \\ =2\int \frac{1}{7+t^2}$ d$t+\frac{2t}{7+t^2}-2\int \frac{1}{7+t^2}$ d$t$ (using IBP)$ \\ =\frac{2 \tan\left (\frac{x}{2} \right ) }{7+\tan^2 \left ( \frac{x}{2} \right )} \\ \\ $ Simplifying this, we get $ \\ \\ \frac{\sin x}{3\cos x +4}+ C$ pretty sure there's a much shorter way haha

InteGrand · Dec 12, 2016

Re: HSC 4U Integration Marathon 2017

$\noindent Let $t\in \mathbb{R}$. Find $\int_{0}^{t} e^{-2x}\sin^{2} x\, \mathrm{d}x$.$

Paradoxica · Dec 13, 2016

Re: HSC 4U Integration Marathon 2017

Let a+b=c where a, b, c are real numbers. Evaluate

$\int \tan{(\text{a}x)}\tan{(\text{b}x)}\tan{(\text{c}x)}\,\text{d}x$

calamebe · Dec 13, 2016

Re: HSC 4U Integration Marathon 2017

InteGrand said:
$\noindent Let $t\in \mathbb{R}$. Find $\int_{0}^{t} e^{-2x}\sin^{2} x\, \mathrm{d}x$.$

www.imgur.com/nGjEjX8

Paradoxica said:
Let a+b=c where a, b, c are real numbers. Evaluate

$\int \tan{(\text{a}x)}\tan{(\text{b}x)}\tan{(\text{c}x)}\,\text{d}x$

www.imgur.com/O3sZNpH

EDIT: That final c should be swapped with a k, but I can't be bothered to edit it and upload another photo.

juantheron · Dec 23, 2016

Re: HSC 4U Integration Marathon 2017

$Evaluation of $\int \frac{x\cos x+1}{\sqrt{2x^3 \cdot e^{\sin x}+x^2}}dx$

$Evaluation of $\int^{\infty}_{-\infty}\frac{\ln(1+16x^2)}{1+25x^2}dx$

stupid_girl · Dec 27, 2016

Re: HSC 4U Integration Marathon 2017

I hope this one isn't too bad.

$\int_{0}^{\pi/4} \frac{log_{2}(cos^{sin 2x}x)}{\pi(3+cos4x)}$

integral95 · Dec 27, 2016

Re: HSC 4U Integration Marathon 2017

stupid_girl said:
I hope this one isn't too bad.

$\int_{0}^{\pi/4} \frac{log_{2}(cos^{sin 2x}x)}{\pi(3+cos4x)}$

A bit tedious though, so after using log laws, u = cos2x, u = tan theta, f(x) = f(a-x), I got -1/64. is that right?

Paradoxica · Dec 27, 2016

Re: HSC 4U Integration Marathon 2017

integral95 said:
A bit tedious though, so after using log laws, u = cos2x, u = tan theta, f(x) = f(a-x), I got -1/64. is that right?

By inspection, you are correct.

HSC 2017 MX2 Integration Marathon (archive) (3 Viewers)

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