HSC 2016 MX2 Integration Marathon (archive) (3 Viewers)

InteGrand · Jul 9, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
Was there any hint towards this step?

Maybe he was thinking of difference of two squares?

leehuan · Jul 9, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
Maybe he was thinking of difference of two squares?

Ah true. But it wouldn't have hit me that placing the brackets around the (sinx + 1) would result in convenience though. Unless it was a lucky gamble

jathu123 · Jul 9, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$\int \frac{\sqrt{1+x^2}}{x} $d$x$

$I =\int \frac{\sqrt{1+x^2}}{x} $d$x \\ $ let $ u=\sqrt{1+x^2} \Rightarrow $ d$x=\frac{\sqrt{1+x^2}}{x}$ d$u \\ I = \int \frac{u^2}{u^2-1}$ d$u$ $ = \int 1+\frac{1}{u^2-1}$ d$u \\ $ let $\frac{A}{u+1}+\frac{B}{u-1}=\frac{1}{u^2-1} \\ $ by expanding and equating the terms, we get $ B-A=1, A+B=0 \\\therefore A= -\frac{1}{2}, B=\frac{1}{2} \\ I= \int 1 $ d$u+\frac{1}{2}\int (\frac{1}{u-1}-\frac{1}{u+1})$ d$u \\ I=u+\frac{1}{2}(\ln (u-1)-\ln (u+1)) \\ \\ I =\sqrt{1+x^2}+\frac{1}{2}\ln\left ( \frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1} \right )+C$

leehuan · Jul 9, 2016

Re: MX2 2016 Integration Marathon

jathu123 said:
$I =\int \frac{\sqrt{1+x^2}}{x} $d$x \\ $ let $ u=\sqrt{1+x^2} \Rightarrow $ d$x=\frac{\sqrt{1+x^2}}{x}$ d$u \\ I = \int \frac{u^2}{u^2-1}$ d$u$ $ = \int 1+\frac{1}{u^2-1}$ d$u \\ $ let $\frac{A}{u+1}+\frac{B}{u-1}=\frac{1}{u^2-1} \\ $ by expanding and equating the terms, we get $ B-A=1, A+B=0 \\\therefore A= -\frac{1}{2}, B=\frac{1}{2} \\ I= \int 1 $ d$u+\frac{1}{2}\int (\frac{1}{u-1}-\frac{1}{u+1})$ d$u \\ I=u+\frac{1}{2}(\ln (u-1)-\ln (u+1)) \\ \\ I =\sqrt{1+x^2}+\frac{1}{2}\ln\left ( \frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1} \right )+C$

May be worth interest, ways to avoid specifically "equating" in partial fractions can be found here:
https://www.maths.unsw.edu.au/sites/default/files/hsc_tips_and_tricks_open_day_booklet_2015.pdf

jathu123 · Jul 9, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
May be worth interest, ways to avoid specifically "equating" in partial fractions can be found here:
https://www.maths.unsw.edu.au/sites/default/files/hsc_tips_and_tricks_open_day_booklet_2015.pdf

Woah that's actually pretty helpful, thanks

Drsoccerball · Jul 9, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
Was there any hint towards this step?

Yes as if you didn't notice the hint. It's called "YOLO."

Drsoccerball · Jul 9, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$I = \int\frac{1 + a x^{t}}{x+bx^{2t+1}} $d$x \quad \forall \, t \in \mathbb{N} \\$

$$ Let $ x^t = u$

$I = \frac{1}{t} \int{\frac{1+au}{u(a+bu^2)}}du$

$$ By applying partial fractions we notice that $ I = \frac{1}{t}\int{(\frac{1}{u} + \frac{a}{1+bu^2} - \frac{bu}{1+bu^2})}du$

$I = ln|x| -\frac{1}{2t} \ln{|1+bx^{2t}|} + \frac{a}{t}\tan^{-1}(bx^t) + C$

leehuan · Jul 9, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
Yes as if you didn't notice the hint. It's called "YOLO."

...

InteGrand · Jul 9, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
...

'Tis an advanced form of 'inspection'.

Drsoccerball · Jul 9, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$I = \int_0^a \frac{\log(1+ax)}{1+x^2} $d$x$$

$$ Let $ x = \tan{(\theta)}$

$\therefore I = \int_0^{\tan^{-1}a}{\log{(1+a\tan{\theta})}d\theta}$

$$ Using $ \int_0^a{f(x)}dx = \int_0^a{f(a-x)}dx$

$I = \int_0^{\tan^{-1}a}{\log{(\frac{1+a^2}{1+a\tan{\theta}})}d\theta}$

$2I = \int_0^{\tan^{-1}a}{\log{(1+a^2)}}d\theta$

$I = \frac{1}{2} \cdot \log{(1+a^2)} \tan^{-1}(a)$

Drsoccerball · Jul 9, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$I = \int \frac{\sqrt{1-x^2}}{2x^4-x^2} $d$x$

$$ Let x = \cos{\theta}$

$I = \int{\tan^2{\theta} \frac{1}{1 - 2\cos^2{\theta}}}d\theta$

$$ Dividing by $ \cos^2{x}$

$I = \int{\frac{\tan^2{\theta}\sec^2{\theta}}{(\tan^2{ \theta } - 1)}}d\theta$

$u = \tan{\theta}$

$I = \int{1 + \frac{1}{2}(\frac{1}{u - 1} - \frac{1}{u + 1})}du$

$I = \tan{\theta} + \frac{1}{2} \ln{|\frac{\tan{\theta - 1}}{\tan{\theta} + 1}|} + C$

$I =\sqrt{\frac{1}{x^2} - 1} + \frac{1}{2} \ln{|\frac{\sqrt{\frac{1}{x^2} - 1} - 1}{\sqrt{\frac{1}{x^2} - 1} + 1}|} + C$

Paradoxica · Jul 9, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
$$ Let $ x = \tan{(\theta)}$

$\therefore I = \int_0^{\tan^{-1}a}{\log{(1+a\tan{\theta})}d\theta}$

$$ Using $ \int_0^a{f(x)}dx = \int_0^a{f(a-x)}dx$

$I = \int_0^{\tan^{-1}a}{\log{(\frac{1+a^2}{1+a\tan{\theta}})}d\theta}$

$2I = \int_0^{\tan^{-1}a}{\log{(1+a^2)}}d\theta$

$I = \frac{1}{2} \cdot \log{(1+a^2)} \tan^{-1}(a)$

Since you're going for the rational inverse mapping you may as well substitute

$x \rightarrow \frac{a-x}{1+ax}$

from the very start. Good work nonetheless.

jathu123 · Jul 9, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$\int \frac{\cos{2x}}{\cot^4{x}}$d$x$

$I= \int \tan ^4x\cos 2x $ d$x \\ $ applying IBP; $ u=\tan ^4x \Rightarrow $ d$u=4\tan ^3x\sec ^2x, \\ $ d$v=\cos 2x \Rightarrow v=\sin x\cos x\\ \\I=\tan ^4x\sin x\cos x-4\int \tan ^3x\sec ^2x \sin x \cos x$ d$x \\ = \sin ^2x\tan ^3x-4\int \tan ^4x$ d$x \\ = \sin ^2x\tan ^3x-4\int( \tan ^2x\sec ^2x - \tan ^2x)$ d$x \\ =\sin ^2x\tan ^3x-4\int (\tan ^2x\sec ^2x - \sec ^2x + 1) $ d$x \\ = \sin ^2x\tan ^3x-\frac{4}{3}\tan^3 x+4\tan x - 4x+C$

Drsoccerball · Jul 9, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$\int \left(\frac{\tan^{-1}{x}}{x-\tan^{-1}{x}}\right)^2 $d$x$

Since this isn't a very common type of questions ill include any thoughts I had. ( Excuse any mistakes pls).

$I noticed that the denominator had a square in it and thus may have something to do with the quotient rule. However the form is too complex to easily inspect so we make a substitution.$

$$ Let $ x = \tan{\theta}$

$I = \int{\frac{\theta^2}{(\sin{\theta} - \theta \cos{\theta})^2}}d\theta$

$\frac{\theta^2}{(\sin{\theta} - \theta \cos{\theta})^2} = \frac{d}{dx} (\frac{f(x)}{\sin{\theta} - \theta \cos{\theta}})$

$$ By expansion the only logical option is either $ f(x) = -\theta\sin{\theta} \ , \ f'(x) = -\theta \cos{\theta} \ , \ f(x) = \frac{-\theta}{\sin{\theta}}.$

$$ By inspection $ f'(x) = -\theta \cos{\theta}$

$I =\frac{\theta\sin{\theta} + \cos{\theta}}{\theta \cos{\theta} - \sin{\theta}} + C$

$$ Divide top and bottom by $ \cos{x}.$

$I = \frac{x\tan^{-1}x + 1}{\tan^{-1}x-x} + C$

juantheron · Jul 10, 2016

Re: MX2 2016 Integration Marathon

$Let $I = \int\frac{\csc^2 x-2005}{\cos^{2005}x}dx = \int \frac{\csc^{2}x}{\cos^{2005}x}dx-\int\frac{2005}{\cos^{2005}}dx$

$Using Integration by parts for $(1)$ one$

$I = -\frac{\cot x}{\cos^{2005}x}+\int 2005 \cos^{-2006}x\cdot \sin x\cdot \cot xdx-\int\frac{2005}{\cos^{2005}x}dx$

$I = -\frac{\cot x}{\cos^{2005}x}+\int \frac{2005}{\cos^{2005}x}dx-\int\frac{2005}{\cos^{2005}x}dx$

$I = \int\frac{\csc^2 x-2005}{\cos^{2005}x}dx = -\frac{\cot x}{\cos^{2005}x}+\mathcal{C}$

RachelGreen · Jul 13, 2016

Re: MX2 2016 Integration Marathon

Hey guys, I got a hard one for you people.

Integrate limits (0 to 2)
[ln(x+2))/x^2+4 .dx

RachelGreen · Jul 13, 2016

Re: MX2 2016 Integration Marathon

Screen shot 2016-07-13 at 1.58.29 PM.png

Here's the question folks

RachelGreen · Jul 13, 2016

Re: MX2 2016 Integration Marathon

Oh and the answer is 3pi/16 x ln2

Paradoxica · Jul 13, 2016

Re: MX2 2016 Integration Marathon

RachelGreen said:
Hey guys, I got a hard one for you people.
...

See here:

http://community.boredofstudies.org...2/350501/impossible-question.html#post7157788

http://community.boredofstudies.org...2/350501/impossible-question.html#post7157764

Set a=2 and your question falls out.

Drsoccerball · Jul 21, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$\int^b_a \dfrac{dx}{\sqrt{(e^x-e^a)(e^b-e^x)}} \quad 0<a<b$

I got something like (assuming no mistakes i'm pretty sure my method was correct). :

$I = \frac{4(b-a)}{(e^b+e^a) \cdot \sqrt{(e^b+e^a)^2 -(b-a)^2}} + C$

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