HSC 2016 MX2 Integration Marathon (archive) (1 Viewer)

InteGrand · Jan 7, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
I thought about this just now

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Edit: fraction should be +2

Nice! The last integral can be quickly dealt with using the auxiliary angle method to transform the denominator into a single sinusoid or cosinusoid. Then have fun transforming everything back in terms of x (something you'd probably feel is the last thing you'd want to do after such a long integral haha).

InteGrand · Jan 7, 2016

Re: MX2 2016 Integration Marathon

Good news, LaTeX appears to be working here again. Test: $x$

Paradoxica · Jan 7, 2016

Re: MX2 2016 Integration Marathon

glittergal96 said:
Yeah, there are a few different ways / times to do it. I don't think any of them will be significantly faster than all of the others though, its pretty much preference.

If it involves long division, for me, that's an automatic no. I avoid long division at all costs.

Here's my method:

$$\noindent$ I_1 = \int \frac{\sqrt{(u+1)^2+1}}{u+2} $d$u = \int \frac{\sqrt{v^2+1}}{v+1} $d$v \\I_1 = \int \frac{v^2-1+2}{(v+1)\sqrt{v^2+1}}$d$v = \int \frac{v-1}{\sqrt{v^2+1}} + \frac{2}{(v+1)\sqrt{v^2+1}} $d$v \\I_1 = \sqrt{v^2+1} - \log{(v+\sqrt{v^2+1})} + \sqrt{2}I_2 \\I_2 = \int \frac{\sqrt{2}}{(v+1)\sqrt{v^2+1}}$d$v\; : v = \tan{\theta} \; , \; $d$v = \sec^2\theta $d$\theta \\I_2=\int \frac{\sqrt{2}\sec^2{\theta} \text{d}\theta}{(1+\tan{\theta})\sec{\theta}}=\int\frac{\sqrt{2}\mathrm{d}\theta}{\sin{\theta} +\cos{\theta}}=\int \frac{\text{d}\theta}{\sin{(\theta + \frac{\pi}{4}})} \\I_2 = -\log\left(\csc \left(\theta+\frac{\pi}{4}\right) + \cot \left(\theta+\frac{\pi}{4}\right)\right) =-\log\left(\frac{\sqrt{2}}{\sin\theta +\cos\theta}+\frac{1-\tan\theta}{1+\tan\theta}\right) \\I_2=-\log\left(\frac{\sqrt{2}\sqrt{v^2+1}}{1+v}+\frac{1-v}{1+v}\right) = \log(1+v)-\log(\sqrt{2}\sqrt{v^2+1}-v+1)$

$$\noindent$I_1 = \sqrt{v^2+1} - \log{v+\sqrt{v^2+1}} + \sqrt{2}\log(1+v)-\sqrt{2}\log(\sqrt{2}\sqrt{v^2+1}-v+1)\\ I_1 = \sqrt{x^2 + 2x + 4 + \frac{2}{x} + \frac{1}{x^2}} - \log{\left (x+1+\frac{1}{x}+\sqrt{x^2 + 2x + 4 + \frac{2}{x} + \frac{1}{x^2}}\right )} + \sqrt{2}\log\left(x+2+\frac{1}{x}\right)-\sqrt{2}\log\left(\sqrt{2}\sqrt{x^2 + 2x + 4 + \frac{2}{x} + \frac{1}{x^2}}-x-\frac{1}{x}\right) \\ I_1 = \frac{\sqrt{x^4 + 2x^3 + 4x^2 + 2x +1}}{x} - \log{\left(x^2 + x + 1 + \sqrt{x^4 + 2x^3 + 4x^2 + 2x +1}\right)} + \log{x} + 2\sqrt{2}\log{|x+1|}-\sqrt{2}\log{\left(\sqrt{2}\sqrt{x^4 + 2x^3 + 4x^2 + 2x +1}-(x^2+1)\right)}+C\\$Where $I_1$ was originally $\int \frac{(x-1)\sqrt{x^4+2x^3+4x^2+2x+1}}{x^2(x+1)}$d$x$

Paradoxica · Jan 7, 2016

Re: MX2 2016 Integration Marathon

$\noindent$\mathbf{NEXT QUESTION}\\ \int\frac{\sin(2x)}{1-\sin^{3}(x)}$d$x\\$There is the brute force method, and then are the cleverer brute force methods$

leehuan · Jan 7, 2016

Re: MX2 2016 Integration Marathon

Just gonna spoil the brute method:

Expand out the double angle, u=sin(x), partial fractions.

Paradoxica · Jan 7, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
Just gonna spoil the brute method:

Expand out the double angle, u=sin(x), partial fractions.

$\noindent And even in the first step towards brute forcing, there is a clever shortcut you can employ to avoid the standard process for decomposing algebraic fractions.$

omegadot · Jan 7, 2016

Re: MX2 2016 Integration Marathon

$\noindent Going back to Paradoxica's 4th integral, once we get to$\\I = \int \frac{\sqrt{u^2 + 2u + 2}}{u + 2} \, du = \int \frac{\sqrt{v^2 + 1}}{v + 1} \, dv\\$the following substitution (a so-called Euler substitution) can be used:$

$\sqrt{v^2 + 1} = v + t \Rightarrow v = \frac{1 - t^2}{2t}, dv = -\frac{1 + t^2}{2t^2} \, dt.$

$$\noindent So grinding this out we have$\\\begin{align*}I &= \int \frac{(t^2 + 1)^2}{2t^2 (t^2 - 2t - 1)} \, dt\\&= \int \left [\frac{1}{2} + \frac{1}{t} - \frac{1}{2t^2} + \frac{4}{t^2 - 2t - 1} \right ] \, dt\\&= \frac{1}{2} \int dt + \int \frac{dt}{t} - \frac{1}{2} \int \frac{dt}{t^2} + 4 \int \frac{dt}{(t - 1)^2 - 2}\\&= \frac{t}{2} + \ln |t| + \frac{1}{2t} + \sqrt{2} \ln \left |\frac{t - 1 - \sqrt{2}}{t - 1 + \sqrt{2}} \right | + \cal{C}.\end{align*}$

$\noindent After substituting $\\\begin{align*}t &= \sqrt{v^2 + 1} - v\\ &= \sqrt{u^2 + 2u + 2} - (u + 1)\\ &= \sqrt{\left (x + \frac{1}{x} \right )^2 + 2 \left (x + \frac{1}{x} \right ) + 2} - \left (x + \frac{1}{x} \right ) - 1\end{align*}\\$the answer to the initial integral should follow.$

$$\noindent Sorry, Paradoxica, but my approach does involve a polynomial long division.$

Paradoxica · Jan 7, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
Just gonna spoil the brute method:

Expand out the double angle, u=sin(x), partial fractions.

$$\noindent Taking leehuan's suggestion, we proceed as follows:$ \\I = \int \frac{2\sin{x}\cos{x}\text{d}x}{1-\sin^3{x}} = \int \frac{2p\text{d}p}{1-p^3}\\ \frac{3}{2}I = \int \frac{3}{1-p^3}\text{d}p = \int \frac{-p^2 + 2p -1 + p^2 + p +1}{(1-p)(p^2+p+1)}$d$p\\\frac{3}{2}I=\int \frac{(1-p)(p-1)}{(1-p)(p^2+p+1)}+\frac{p^2+p+1}{(1-p)(p^2+p+1)}$d$p\\\frac{3}{2}I = \int \frac{p-1}{p^2+p+1}-\frac{1}{p-1}$d$p = \int \frac{p+\frac{1}{2}}{p^2+p+1}-\frac{\frac{3}{2}}{p^2+p+1}-\frac{1}{p-1} $ d$p\\ \frac{3}{2}I = \frac{1}{2}\log{(p^2+p+1)}-\frac{3}{2}\cdot \frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{2p-1}{\sqrt{3}}\right)-\log(p-1)+C \\ I = \frac{1}{3}\log{(p^2+p+1)}-\frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{2p-1}{\sqrt{3}}\right)-\log(p-1)$

InteGrand · Jan 7, 2016

Re: MX2 2016 Integration Marathon

$\textbf{NEW QUESTION}$

$\noindent Find $\int \frac{1}{x+\sqrt{x^2 +x + 1}}\text{ d}x$.$

Paradoxica · Jan 8, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
$\textbf{NEW QUESTION}$

$\noindent Find $\int \frac{1}{x+\sqrt{x^2 +x + 1}}\text{ d}x$.$

$\noindent Rationalise the denominator to yield $I_1 = \int \frac{\sqrt{x^2+x+1}-x}{x+1}$d$x\\I_1 = 2I_2 -x + \log{(x+1)}\\I_2=2\int \frac{\sqrt{(2x+1)^2+3}}{2x+1+1}$d$x=\int \frac{\sqrt{b^2+3}}{b+1}$d$b\\I_2$ can now be completed in exactly the same manner as my most recent difficult question, again by rationalising the numerator to split off a part which yields a surd and inverse hyperbolic sine, and another part which can be dealt with by a trigonometric substitution. Even the auxiliary transfomation will be nice due to the presence of $\sqrt{3}$ throughout the integral.$

KingOfActing · Jan 8, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
$\textbf{NEW QUESTION}$

$\noindent Find $\int \frac{1}{x+\sqrt{x^2 +x + 1}}\text{ d}x$.$

Paradoxica said:
$\noindent Rationalise the denominator to yield $I_1 = \int \frac{\sqrt{x^2+x+1}-x}{x+1}$d$x\\I_1 = 2I_2 -x + \log{(x+1)}\\I_2=2\int \frac{\sqrt{(2x+1)^2+3}}{2x+1+1}$d$x=\int \frac{\sqrt{b^2+3}}{b+1}$d$b\\I_2$ can now be completed in exactly the same manner as my most recent difficult question, again by rationalising the numerator to split off a part which yields a surd and inverse hyperbolic sine, and another part which can be dealt with by a trigonometric substitution. Even the auxiliary transfomation will be nice due to the presence of $\sqrt{3}$ throughout the integral.$

I did it a kinda similar way, but with just a lot of substitutions (I like substitutions okay).

$\begin{align*}\text{Let } x + \frac{1}{2} = \frac{\sqrt3}{2}\sinh{u}\\\int \frac{1}{x+\sqrt{x^2+x+1}}\,dx &= \int \frac{\sqrt{3}\cosh{u}}{\sqrt3 e^u-1}\,du\\\text{Now, let }u = \ln{z}\\&=\frac{1}{2} \int \frac{z^2+1}{z^2(z-\frac{1}{\sqrt3})}\,dz\\\text{Now, by partial fraction decomposition: }\\&=\frac{1}{2}\left(\int\frac{-3}{z}\,dz - \int\frac{\sqrt3}{z^2}\,dz + \int\frac{4}{z-\frac{1}{3}}\,dz\right)\\&= \frac{-3}{2}\ln|z| + \frac{\sqrt3}{2z} + 2\ln|z - \frac{1}{\sqrt3}|+ C\\\text{Now, just back substituting twice: }\\&=\sqrt{x^2+x+1} - x + 2\ln|x+\sqrt{x^2+x+1}| - \frac{3}{2}\sinh^{-1}\left(\frac{2x+1}{\sqrt{3}}\right) + C\end{align*}$

omegadot · Jan 8, 2016

Re: MX2 2016 Integration Marathon

I didn't think the hyperbolic functions and their inverses were part of the standard MX2 course. Am I mistaken?

KingOfActing · Jan 8, 2016

Re: MX2 2016 Integration Marathon

omegadot said:
I didn't think the hyperbolic functions and their inverses were part of the standard MX2 course. Am I mistaken?

We've already gone over them in my 3U class, so I don't know.

InteGrand · Jan 8, 2016

Re: MX2 2016 Integration Marathon

omegadot said:
I didn't think the hyperbolic functions and their inverses were part of the standard MX2 course. Am I mistaken?

You're not mistaken. Some teachers might teach it as an 'extension'.

leehuan · Jan 8, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$$\noindent Taking leehuan's suggestion, we proceed as follows:$ \\I = \int \frac{2\sin{x}\cos{x}\text{d}x}{1-\sin^3{x}} = \int \frac{2p\text{d}p}{1-p^3}\\ \frac{3}{2}I = \int \frac{3}{1-p^3}\text{d}p = \int \frac{-p^2 + 2p -1 + p^2 + p +1}{(1-p)(p^2+p+1)}$d$p\\\frac{3}{2}I=\int \frac{(1-p)(p-1)}{(1-p)(p^2+p+1)}+\frac{p^2+p+1}{(1-p)(p^2+p+1)}$d$p\\\frac{3}{2}I = \int \frac{p-1}{p^2+p+1}-\frac{1}{p-1}$d$p = \int \frac{p+\frac{1}{2}}{p^2+p+1}-\frac{\frac{3}{2}}{p^2+p+1}-\frac{1}{p-1} $ d$p\\ \frac{3}{2}I = \frac{1}{2}\log{(p^2+p+1)}-\frac{3}{2}\cdot \frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{2p-1}{\sqrt{3}}\right)-\log(p-1)+C \\ I = \frac{1}{3}\log{(p^2+p+1)}-\frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{2p-1}{\sqrt{3}}\right)-\log(p-1)$

Ah nice, it's still partial fractions but it's partial fractions by inspection. I can only do that with linear denominators.

KingOfActing said:
We've already gone over them in my 3U class, so I don't know.

Stop using hyperbolic functions.

braintic · Jan 8, 2016

Re: MX2 2016 Integration Marathon

omegadot said:
I didn't think the hyperbolic functions and their inverses were part of the standard MX2 course. Am I mistaken?

They have been asked indirectly in older HSC exams, where the functions C(x) and S(x) were defined in terms of exponentials, and the various properties required to be derived.

leehuan · Jan 8, 2016

Re: MX2 2016 Integration Marathon

braintic said:
They have been asked indirectly in older HSC exams, where the functions C(x) and S(x) were defined in terms of exponentials, and the various properties required to be derived.

Whilst I picked up the hint, they will forever remain indirect questions.

DatAtarLyfe · Jan 8, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
You're not mistaken. Some teachers might teach it as an 'extension'.

IIRC, there is a question in the 3u cambridge under inverse trig that works through hyperbolic functions, not sure if it's extension...

InteGrand · Jan 8, 2016

Re: MX2 2016 Integration Marathon

DatAtarLyfe said:
IIRC, there is a question in the 3u cambridge under inverse trig that works through hyperbolic functions, not sure if it's extension...

Yeah, there is definitely a question there about their derivatives' properties, but they don't mention that these functions are the hyperbolic functions. HSC students aren't expected to know the properties of these functions, or even what these functions are. They'd be indirect questions if asked, as leehuan said. Such questions can be done without any knowledge of the hyperbolic functions.

DatAtarLyfe · Jan 8, 2016

Re: MX2 2016 Integration Marathon

oh, well my maths teacher taught us as well, he said "it's good to have knowledge of them so that you can approach questions more uniquely"
Idk he was probs high

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