HSC 2016 MX2 Integration Marathon (archive) (2 Viewers)

leehuan · Jan 6, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
The method for the last integral is obvious if you know what a symmetric substitution is.

I can't find an article on this

leehuan · Jan 6, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
Maybe he means a "self-inverse" substitution. Is that what you meant, Paradoxica?

Could still do with an article D:

Paradoxica · Jan 6, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
I can't find an article on this

$$Ah, yes, there's no official consensus on it, but a symmetric substitution is one where you have a rational function written in the form $P(x) \pm P\left(\frac{1}{x}\right)$

omegadot · Jan 6, 2016

Re: MX2 2016 Integration Marathon

$\noindent I am sorry Paradoxia but I don't quite have a simple three-liner solution to this one. Your solution is none-the-less very impressive! To address ze-'s concern about the level of difficult of questions asked here, let me rewrite my question in a form it would most likely appear on a MX2 paper. In doing so it allows me to take you through the method of solution I used.$

$$\noindent \textbf{Question \`{a} la HSC MX2 style}$

$$\noindent Let $I = \int^1_0 \left (\frac{2}{x + 1} - x - 1 \right )^{2n} \, dx$ where $n \in \mathbb{N}$.$

$$\noindent (a) By using the substitution $u = \frac{2}{x + 1} - 1$, show that the integral can be rewritten as$\\I = \int^1_0 \frac{2}{(u + 1)^2} \left (\frac{2}{u + 1} - u - 1 \right )^{2n} \, du.$

$$\noindent (b) By adding the two forms for the integral $I$ together, show that$\\I = \frac{1}{2} \int^1_0 \left (\frac{2}{(x + 1)^2} + 1 \right ) \left (\frac{2}{x + 1} - x - 1 \right )^{2n} \, dx.$

$$\noindent (c) Now using the substitution $u = \frac{2}{x + 1} - x - 1$ in the integral given in (b), deduce that $I = \frac{1}{2n + 1}.$

$$\noindent \textbf{Bonus Question}$

$$\noindent By following the same approach as used in the above question, show that$\\\begin{align*}\int^{a+ b}_{a + 1} \left (\frac{b}{x - a} + a - x \right )^{2n} dx &= \frac{1}{2(2n + 1)} \left [(b - 1)^{2n + 1} - (1 - b)^{2n + 1} \right ], \quad n \in \mathbb{N}.\end{align*}$

leehuan · Jan 6, 2016

Re: MX2 2016 Integration Marathon

omegadot said:
$\noindent I am sorry Paradoxia but I don't quite have a simple three-liner solution to this one. Your solution is none-the-less very impressive! To address ze-'s concern about the level of difficult of questions asked here, let me rewrite my question in a form it would most likely appear on a MX2 paper. In doing so it allows me to take you through the method of solution I used.$

$$\noindent \textbf{Question \`{a} la HSC MX2 style}$

$$\noindent Let $I = \int^1_0 \left (\frac{2}{x + 1} - x - 1 \right )^{2n} \, dx$ where $n \in \mathbb{N}$.$

$$\noindent (a) By using the substitution $u = \frac{2}{x + 1} - 1$, show that the integral can be rewritten as$\\I = \int^1_0 \frac{2}{(u + 1)^2} \left (\frac{2}{u + 1} - u - 1 \right )^{2n} \, du.$

$$\noindent (b) By adding the two forms for the integral $I$ together, show that$\\I = \frac{1}{2} \int^1_0 \left (\frac{2}{(x + 1)^2} + 1 \right ) \left (\frac{2}{x + 1} - x - 1 \right )^{2n} \, dx.$

$$\noindent (c) Now using the substitution $u = \frac{2}{x + 1} - x - 1$ in the integral given in (b), deduce that $I = \frac{1}{2n + 1}.$

$$\noindent \textbf{Bonus Question}$

$$\noindent By following the same approach as used in the above question, show that$\\\begin{align*}\int^{a+ b}_{a + 1} \left (\frac{b}{x - a} + a - x \right )^{2n} dx &= \frac{1}{2(2n + 1)} \left [(b - 1)^{2n + 1} - (1 - b)^{2n + 1} \right ], \quad n \in \mathbb{N}.\end{align*}$

For the 2016ers... you will be guided through this in the HSC...

Paradoxica · Jan 6, 2016

Re: MX2 2016 Integration Marathon

omegadot said:
$\noindent I am sorry Paradoxia but I don't quite have a simple three-liner solution to this one. Your solution is none-the-less very impressive! To address ze-'s concern about the level of difficult of questions asked here, let me rewrite my question in a form it would most likely appear on a MX2 paper. In doing so it allows me to take you through the method of solution I used.$

$$\noindent \textbf{Question \`{a} la HSC MX2 style}$

$$\noindent Let $I = \int^1_0 \left (\frac{2}{x + 1} - x - 1 \right )^{2n} \, dx$ where $n \in \mathbb{N}$.$

$$\noindent (a) By using the substitution $u = \frac{2}{x + 1} - 1$, show that the integral can be rewritten as$\\I = \int^1_0 \frac{2}{(u + 1)^2} \left (\frac{2}{u + 1} - u - 1 \right )^{2n} \, du.$

$$\noindent (b) By adding the two forms for the integral $I$ together, show that$\\I = \frac{1}{2} \int^1_0 \left (\frac{2}{(x + 1)^2} + 1 \right ) \left (\frac{2}{x + 1} - x - 1 \right )^{2n} \, dx.$

$$\noindent (c) Now using the substitution $u = \frac{2}{x + 1} - x - 1$ in the integral given in (b), deduce that $I = \frac{1}{2n + 1}.$

$$\noindent \textbf{Bonus Question}$

$$\noindent By following the same approach as used in the above question, show that$\\\begin{align*}\int^{a+ b}_{a + 1} \left (\frac{b}{x - a} + a - x \right )^{2n} dx &= \frac{1}{2(2n + 1)} \left [(b - 1)^{2n + 1} - (1 - b)^{2n + 1} \right ], \quad n \in \mathbb{N}.\end{align*}$

$$Just saying, that answer can be simplified to $\frac{(b-1)^{2n+1}}{2n+1}$

InteGrand · Jan 6, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$$Just saying, that answer can be simplified to $\frac{(b-1)^{2n+1}}{2n+1}$

$\noindent I think it was a typo by omegadot. The final answer should clearly depend on $a$, so one of those terms in the bracket should probably have an $a$ in it.$

glittergal96 · Jan 6, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
$\noindent I think it was a typo by omegadot. The final answer should clearly depend on $a$, so one of those terms in the bracket should probably have an $a$ in it.$

It shouldn't depend on a, the a drops out after the translation y=x-a.

InteGrand · Jan 6, 2016

Re: MX2 2016 Integration Marathon

glittergal96 said:
It shouldn't depend on a, the a drops out after the translation y=x-a.

Ah right, of course. My bad.

Paradoxica · Jan 6, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
$Meanwhile, for even exponents, we can similarly show by considering binomial expansion that the power of $x$ in the general term is always even (whereas before it was always odd). This means there is no $x^{-1}$ term, so all terms integrate to give rational values in the end. So the answer is always \emph{rational}, so the pattern is different to the odd exponent case. I haven't investigated whether there's a pattern for the answers in the odd exponents case, but I wouldn't be surprised if there is.$

There is a "pattern", but due to the irrationality of the integral, it requires binomial co-efficients and tedious use of common-denominators.

hedgehog_7 · Jan 7, 2016

Re: MX2 2016 Integration Marathon

Given P ( a cos theta , b sin theta ) and Q ( a cos phi , b sin phi ) lie on the ellipse x^2/a^2 + y^2/b^2 = 1

IF PQ subtends a right angle at (a,0) show that ( tan theta/2 ) X ( tan phi/2 ) = -b^2/a^2

hedgehog_7 · Jan 7, 2016

Re: MX2 2016 Integration Marathon

also sorry for double post i think this conics stuff is really messing with me, this is some stupid trig identity thing with this conics that isnt really clicking with me:

How does b (sin theta - sin phi ) / a (cos theta - cos phi) = b/a * ( 2 sin ( (theta - phi) /2 ) ) ) * cos ( theta + phi /2 ) ) / - 2 sin ( (theta - phi) /2 ) ) sin (theta + phi /2 )
Sorry i might have really butchered this in typing it out. I couldnt upload my photo because file size was too big? :L page 89 conics cambridge 4u . Any help would be awesome! thanks

Paradoxica · Jan 7, 2016

Re: MX2 2016 Integration Marathon

hedgehog_7 said:
also sorry for double post i think this conics stuff is really messing with me, this is some stupid trig identity thing with this conics that isnt really clicking with me:

$$How does $\frac{b(\sin\theta - \sin\phi )}{a (\cos\theta -\cos\phi)} = \frac{b}{a}\frac{2\sin\frac{\theta - \phi}{2}\cos\frac{\theta + \phi}{2}}{-2\sin\frac{\theta -\phi}{2}\sin\frac{\theta +\phi}{2}}$
Sorry i might have really butchered this in typing it out. I couldnt upload my photo because file size was too big? :L page 89 conics cambridge 4u . Any help would be awesome! thanks

Wrong thread, put this on the standard 4U marathon, or make a new thread for this question.
Use the sum to product identity.

Paradoxica said:
$4. \int \frac{(x-1)\sqrt{x^4+2x^3+4x^2+2x+1}}{x^2(x+1)}$d$x$

$$\noindent To aid the completion of this integral, your hint is the symmetric substitution $d = x+\frac{1}{x}$

leehuan · Jan 7, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
Wrong thread, put this on the standard 4U marathon, or make a new thread for this question.
Use the sum to product identity.

$$\noindent To aid the completion of this integral, your hint is the symmetric substitution $d = x+\frac{1}{x}$

I eventually deduced the substitution, but lack of experience with using it meant I couldn't make any progress. Will wait for a solution to use as a learning tool.

(but not a rushed Drsoccerball solution lol)

glittergal96 · Jan 7, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
Bloody hell that Recurrence formula is a bitch.

$$\noindent Replace $x+1$ with $x$ and shift the border of integration by one unit to obtain:$\\ \int^1_0 \left (\frac{2}{x + 1} - x - 1 \right )^{2014}$d$x \equiv \int_1^2 \left ( \frac{2}{x} -x \right )^{2014}$d$x\\$Claim: $P(n) : \int_1^2 \left ( \frac{2}{x} -x \right )^{2n}$d$x \equiv \frac{1}{2n+1}\; \forall n \in \mathbb{N} \\$Proof: $P(0)$ is trivial. For $n=1$, we have: $\\ \int_1^2 \left (\frac{2}{x}-x \right )^2 $d$x = \int_1^2 \frac{4}{x^2} - 4 + x^2 $d$x =\left [ -\frac{4}{x} -4x + \frac{x^3}{3} \right ]_1^2 \\ = -2-8+\frac{8}{3}+4+4-\frac{1}{3}=\frac{1}{3} \Rightarrow P(1)$ is true.$\\$Assume $P(n)$ is true: $ \rightarrow \int_1^2 \left ( \frac{2}{x} -x \right )^{2n}$d$x = \frac{1}{2n+1} \\I_1 = \int_1^2 \left ( \frac{2}{x} -x \right )^{2n+2}$d$x = \left[x\left(\frac{2}{x}-x\right)^{2n+2}\right]_1^2-(2n+2)\int_1^2 \left(\frac{2}{x}-x\right)^{2n+1}x\left(-\frac{2}{x^2}-1\right)$d$x$

$$\noindent$I_1 = 2\left(\frac{2}{2}-2\right)^{2n+2}-1\left(\frac{2}{1}-1\right)^{2n+2}+(2n+2)\int_1^2\left(\frac{2}{x}-x\right)^{2n+1}\left(\frac{2}{x}-x+2x\right)$d$x\\I_1 = 1+(2n+2)\int_1^2\left(\frac{2}{x}-x\right)^{2n+2}$d$x+(2n+2)\int_1^22x\left(\frac{2}{x}-x\right)^{2n+1}$d$x\\I_1 = 1+(2n+2)I_1 +(2n+2)\int_1^2\left(\frac{2}{x}-x\right)^{2n+1}$d$(x^2)\\I_2 = \int_1^2\left(\frac{2}{x}-x\right)^{2n+1}$d$(x^2) = \left[x^2\left(\frac{2}{x}-x\right)^{2n+1}\right]_1^2-(2n+1)\int_1^2x^2\left(\frac{2}{x}-x\right)^{2n}\left(-\frac{2}{x^2}-1\right)$d$x\\I_2 = 4\left(\frac{2}{2}-2\right)^{2n+1}-1\left(\frac{2}{1}-1\right)^{2n+1}+(2n+1)\int_1^2x^2\left(\frac{2}{x}-x\right)^{2n}\left(\frac{2}{x^2}+1\right)$d$x\\I_2=-5+(2n+1)I_3\\I_3=\int_1^2 x\left(\frac{2}{x}+x\right)\left(\frac{2}{x}-x\right)^{2n}$d$x=\int_1^2 (x^2+2)\left(\frac{2}{x}-x\right)^{2n}$d$x\\I_3=\int_1^2x^2\left(\frac{2}{x}-x\right)^{2n}\left(\frac{2}{x^2}+1\right)$d$x\\$First evaluation of $I_3:\\$

$$\noindent$I_3=\int_1^2 x\left(\frac{2}{x}+x\right)\left(\frac{2}{x}-x\right)^{2n}$d$x=\int_1^2x\left(\frac{2}{x}-x\right)^{2n}\left(\frac{2}{x}-x+2x\right)$d$x\\I_3=\int_1^2 \left(\frac{2}{x}-x\right)^{2n+1}$d$\left(\frac{x^2}{2}\right)+2\int_1^2x^2\left(\frac{2}{x}-x\right)^{2n}$d$x\\I_3=\left[\frac{x^2}{2}\left(\frac{2}{x}-x\right)^{2n+1}\right]_1^2+\frac{2n+1}{2}\int_1^2x^2\left(\frac{2}{x}-x\right)^{2n}\left(\frac{2}{x^2}+1\right)$d$x+2I_4\\I_3=-\frac{5}{2}+\frac{2n+1}{2}I_3+2I_4\\$Second Evaluation of $I_3:\\I_3=\int_1^2 (x^2+2)\left(\frac{2}{x}-x\right)^{2n}$d$x=2\int_1^2\left(\frac{2}{x}-x\right)^{2n}$d$x+\int_1^2x^2\left(\frac{2}{x}-x\right)^{2n}$d$x\\I_3=\frac{2}{2n+1}+I_4\\2I_3-I_3=\frac{4}{2n+1}+2I_4+\frac{5}{2}-\frac{2n+1}{2}I_3-2I_4\\\frac{2n+3}{2}I_3=\frac{10n+13}{2(2n+1)} \Leftrightarrow I_3 = \frac{10n+13}{(2n+1)(2n+3)}\\I_2=-5+(2n+1)I_3=\frac{10n+13}{2n+3}-5\\I_1=1+(2n+2)I_1+(2n+2)I_2=1+(2n+2)I_1+\frac{(2n+2)(10n+13)}{2n+3}-(10n+10)$

$$\noindent$(2n+1)I_1=\frac{(10n+9)(2n+3)-(2n+2)(10n+13)}{2n+3}\\(2n+1)I_1=\frac{20n^2+48n+27-20n^2-46n-26}{2n+3}=\frac{2n+1}{2n+3}\\\therefore \int_1^2 \left(\frac{2}{x}-x\right)^{2n+2}$d$x=\frac{1}{2n+3}\\\therefore P(n) \Rightarrow P(n+1)\\\therefore$ by the Axiom of Induction, $ P(n)$ is true $\forall n \in \mathbb{N}\\\therefore \int_1^2 \left(\frac{2}{x}-x\right)^{2014}$d$x=\frac{1}{2015}\\\therefore \int_0^1 \left(\frac{2}{x+1}-x-1\right)^{2014}$d$x = \frac{1}{2015}\\\textit{QUOD ERAT-FUCKING-DEMONSTRATUM}$

Definitely don't need any recursion for this integral. Here is my solution to the generalisation where 2 is replaced by b:

$J_n=\int_1^b \left(\frac{b}{x}-x\right)^{2n}\, dx\\ \\ = \frac{1}{2}\int_{b-1}^{1-b}u^{2n}\left(-1+\frac{u}{\sqrt{u^2+4b}}\right)\, du\\ \\ =\int_0^{b-1}u^{2n}\, du\\ = \frac{(b-1)^{2n+1}}{2n+1}\\ \\ \\ $where the second line comes from the substitution$\\ x=-u+\sqrt{u^2+4b}\Rightarrow u=b/x-x\\ \\ $and the third line comes from dropping an odd term in the integrand.$$

Paradoxica · Jan 7, 2016

Re: MX2 2016 Integration Marathon

glittergal96 said:
Definitely don't need any recursion for this integral. Here is my solution to the generalisation where 2 is replaced by b:

$J_n=\int_1^b \left(\frac{b}{x}-x\right)^{2n}\, dx\\ \\ = \frac{1}{2}\int_{b-1}^{1-b}u^{2n}\left(-1+\frac{u}{\sqrt{u^2+4b}}\right)\, du\\ \\ =\int_0^{b-1}u^{2n}\, du\\ = \frac{(b-1)^{2n+1}}{2n+1}\\ \\ \\ $where the second line comes from the substitution$\\ x=-u+\sqrt{u^2+4b}\Rightarrow u=b/x-x\\ \\ $and the third line comes from dropping an odd term in the integrand.$$

Yeah, I'm only used to symmetric substitutions, these ones I haven't learnt to see yet.

Specifically asymmetric substitutions of the rational kind.

glittergal96 · Jan 7, 2016

Re: MX2 2016 Integration Marathon

The motivation is fairly straightforward thankfully: u=b/x-x is a pretty logical thing to try (simplify the ugliest part of the integrand).

Then I just arbitrarily (but consistently) chose one of the solutions to the resulting quadratic equation in x.

Spotting a substitution like that if that term wasn't isolated under the power would be an entirely different matter.

glittergal96 · Jan 7, 2016

Re: MX2 2016 Integration Marathon

Of course, this method also makes it clear why there is more work to do when integrating (k/x-x)^(2n+1).

With this switched parity, the SIMPLER term in the integrand drops out from being odd and the more complicated one survives.

This means that we essentially have to integrate things of the form I_m := x^(2m)/sqrt(x^2+4b) from 0 to b-1.

We can attack this guy with trig/hyp trig (the sinh sub almost immediately gives us a recurrence, but I'm not sure if you will be able to find a nice closed form solution for I_m).

Paradoxica · Jan 7, 2016

Re: MX2 2016 Integration Marathon

glittergal96 said:
Of course, this method also makes it clear why there is more work to do when integrating (k/x-x)^(2n+1).

With this switched parity, the SIMPLER term in the integrand drops out from being odd and the more complicated one survives.

This means that we essentially have to integrate things of the form I_m := x^(2m)/sqrt(x^2+4b) from 0 to b-1.

We can attack this guy with trig/hyp trig (the sinh sub almost immediately gives us a recurrence, but I'm not sure if you will be able to find a nice closed form solution for I_m).

The general solution of the expansion will involve log(2), which means it's not going to be nice. I've already numerically analysed the first few odd powers and the only possibility of a pattern involves tedious combinations of binomial expressions. The co-efficient for log(2), however is quite nice (always an integer) due to the log(2) term being strictly dependent on a single term of the binomial expansion.

Now glittergal, why don't you solve my 4th integral for leehuan?

glittergal96 · Jan 7, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
The general solution of the expansion will involve log(2), which means it's not going to be nice. I've already numerically analysed the first few odd powers and the only possibility of a pattern involves tedious combinations of binomial expressions. The co-efficient for log(2), however is quite nice (always an integer) due to the log(2) term being strictly dependent on a single term of the binomial expansion.

Now glittergal, why don't you solve my 4th integral for leehuan?

Well, I mean the existence of a log term doesn't in itself mean the answer cannot be found in closed form.
Finding a binomial expansion for the integral is immediate, but its not obvious that this expansion does not have a closed form.

Sure, I will have a crack at it later this arvo if no-one else does. There are just a few other things I want to finish first.

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