# Thread: HSC 2017 MX2 Integration Marathon (archive)

1. ## Re: HSC 4U Integration Marathon 2017

Originally Posted by InteGrand
Find

$\int \frac { x^{ \frac { 2 }{ 3 } } }{ \left( 1+x^{ \frac { 3 }{ 5 } } \right) ^{ 2 } }\, \mathrm{d}x$.

(Courtesy of leehuan here: http://community.boredofstudies.org/...ml#post7183192 )
Oh gd.

2. ## Re: HSC 4U Integration Marathon 2017

$\noindent Let a be a real constant with |a| \neq 1. Find \int \frac{1-a^{2}}{1-2a\cos x + a^{2}}\, \mathrm{d}x.$

3. ## Re: HSC 4U Integration Marathon 2017

Originally Posted by InteGrand
$\noindent Let a be a real constant with |a| \neq 1. Find \int \frac{1-a^{2}}{1-2a\cos x + a^{2}}\, \mathrm{d}x.$
$\noindentI= \int \frac{1-a^2}{1-2a \cos x + a^2} dx \\ = (1-a^2)\int\frac{1}{1+a^2-2a(2\cos^2(\frac{x}{2})-1)} dx \\(1-a^2) \int \frac{1}{(1+a)^2-4a \cos^2 \frac{x}{2}} dx\\\\ Let t=\tan \frac{x}{2}\Rightarrow dt=\frac{1}{2}\sec ^2\frac{x}{2} dx\\\therefore dx=2 \cos^2 \frac{x}{2} dt = \frac{2}{1+t^2} dt \\\\ I=2(1-a^2)\int \frac{1}{(1+a)^2-\frac{4a}{1+t^2}}\cdot \frac{1}{1+t^2} dt \\ = 2(1-a^2)\int \frac{1}{(1-a)^2+t^2(1+a)^2} dt \\ = 2(1-a^2)\left (\frac{1}{1-a^2} \tan^{-1}\frac{t(1+a)}{1-a} \right )+C \\ \\ = 2\tan^{-1}\left ( \frac{1+a}{1-a} \tan \frac{x}{2} \right ) + C$

4. ## Re: HSC 4U Integration Marathon 2017

Originally Posted by jathu123
$\noindentI= \int \frac{1-a^2}{1-2a \cos x + a^2} dx \\ = (1-a^2)\int\frac{1}{1+a^2-2a(2\cos^2(\frac{x}{2})-1)} dx \\(1-a^2) \int \frac{1}{(1+a)^2-4a \cos^2 \frac{x}{2}} dx\\\\ Let t=\tan \frac{x}{2}\Rightarrow dt=\frac{1}{2}\sec ^2\frac{x}{2} dx\\\therefore dx=2 \cos^2 \frac{x}{2} dt = \frac{2}{1+t^2} dt \\\\ I=2(1-a^2)\int \frac{1}{(1+a)^2-\frac{4a}{1+t^2}}\cdot \frac{1}{1+t^2} dt \\ = 2(1-a^2)\int \frac{1}{(1-a)^2+t^2(1+a)^2} dt \\ = 2(1-a^2)\left (\frac{1}{1-a^2} \tan^{-1}\frac{t(1+a)}{1-a} \right )+C \\ \\ = 2\tan^{-1}\left ( \frac{1+a}{1-a} \tan \frac{x}{2} \right ) + C$
How do you already know 4u integration wtf

5. ## Re: HSC 4U Integration Marathon 2017

Originally Posted by pikachu975
How do you already know 4u integration wtf
He probably accelerated ahead of himself, how else would he get up to 4U integration?

6. ## Re: HSC 4U Integration Marathon 2017

Originally Posted by pikachu975
How do you already know 4u integration wtf

7. ## Re: HSC 4U Integration Marathon 2017

Originally Posted by eyeseeyou
He probably accelerated ahead of himself, how else would he get up to 4U integration?
Not that easy to self learn the 4u course, unless you have a lot of spare time and really like maths.

8. ## Re: HSC 4U Integration Marathon 2017

Originally Posted by Drsoccerball
lmao no way, I shouldn't be even compared to him

9. ## Re: HSC 4U Integration Marathon 2017

Originally Posted by pikachu975
How do you already know 4u integration wtf
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.

10. ## Re: HSC 4U Integration Marathon 2017

Originally Posted by jathu123
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.
What other topics are you familiar with?

11. ## Re: HSC 4U Integration Marathon 2017

Originally Posted by pikachu975
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.

12. ## Re: HSC 4U Integration Marathon 2017

Originally Posted by jathu123
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.

13. ## Re: HSC 4U Integration Marathon 2017

Originally Posted by jathu123
lmao no way, I shouldn't be even compared to him
Become my student and one day you will best me.

14. ## Re: HSC 4U Integration Marathon 2017

Become my student and one day you will best me.

15. ## Re: HSC 4U Integration Marathon 2017

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

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

16. ## Re: HSC 4U Integration Marathon 2017

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

17. ## 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
\begin{align*}\text{And }\sqrt{x-x^2}&=\sqrt{(1-x)-(1-x)^2}\\ &\text{is positive for }0
$\int_{0}^{\frac{1}{2}}{\cos{(\pi x)}\text{d}x}=-\int_{\frac{1}{2}}^{1}{\cos{(\pi x)}\text{d}x}$
$\text{So }1+\frac{1}{\pi}\int_{0}^{1}{\frac{\cos{(\pi x)}}{\sqrt{x-x^2}}\text{d}x}=1+\frac{0}{\pi}=1$
$\therefore\int_{-1}^{1}{\sin{(\pi |x|)}\sin^{-1}{\sqrt{|x|}}\text{d}x}=1$

18. ## Re: HSC 4U Integration Marathon 2017

Originally Posted by juantheron
$Evaluation of \int^{\frac{\pi}{4}}_{0}(\cos 2x)^{\frac{3}{2}}\cdot \cos xdx$
$\sqrt{2}\sin x = \sin \theta \implies \cos x \, dx =\frac1{\sqrt2}\cos \theta \, d\theta$

\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.}

19. ## Re: HSC 4U Integration Marathon 2017

Let's try that again.

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

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

20. ## 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*}
$\therefore\int_{-1}^{1}{\sin{(\pi|x|)}\sin^{-1}{\sqrt{|x|}}\text{d}x}=1$

21. ## Re: HSC 4U Integration Marathon 2017

Originally Posted by Lugia101101
\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.

22. ## 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*}

23. ## Re: HSC 4U Integration Marathon 2017

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

24. ## Re: HSC 4U Integration Marathon 2017

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

25. ## Re: HSC 4U Integration Marathon 2017

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

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

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