HSC 2012-14 MX2 Integration Marathon (archive) (4 Viewers)

braintic · Jul 6, 2013

Re: MX2 Integration Marathon

Sy123 said:
$For some non-negative integer$ \ \ n \ \ $evaluate$

$\int_0^1 \frac{x^{2n}}{1-x^2} \ dx$

I think there is a problem there, considering the upper limit and the denominator.
And Wolfram says "integral does not converge".

Sy123 · Jul 6, 2013

Re: MX2 Integration Marathon

braintic said:
I think there is a problem there, considering the upper limit and the denominator.
And Wolfram says "integral does not converge".

Ah upper limit is supposed to be 1/2, not 1 sorry about that.

braintic · Jul 6, 2013

Re: MX2 Integration Marathon

Sy123 said:
$For some non-negative integer$ \ \ n \ \ $evaluate$

$\int_0^{1/2} \frac{x^{2n}}{1-x^2} \ dx$

I can only turn this into a recurrence relation leading to (ln3)/2 minus a sum of n terms, each term being 1/[(2k-1).2^k]. Is this correct, and is it possible to sum this series?

RealiseNothing · Jul 6, 2013

Re: MX2 Integration Marathon

Sy123 said:
$For some non-negative integer$ \ \ n \ \ $evaluate$

$\int_0^{1/2} \frac{x^{2n}}{1-x^2} \ dx$

$\frac{x^{2n}}{1-x^2}$

$-(\frac{1-x^{2n}-1}{1-x^2})$

$-(\frac{1-x^{2n}}{1-x^2} - \frac{1}{1-x^2})$

$\frac{1}{1-x^2} - \frac{1-x^{2n}}{1-x^2}$

$\frac{1}{1-x^2} - 1 - x^2 - x^4 - ... - x^{2n-2}$

Then just integrate this which is straight forward.

Sy123 · Jul 6, 2013

Re: MX2 Integration Marathon

RealiseNothing said:
$\frac{x^{2n}}{1-x^2}$

$-(\frac{1-x^{2n}-1}{1-x^2})$

$-(\frac{1-x^{2n}}{1-x^2} - \frac{1}{1-x^2})$

$\frac{1}{1-x^2} - \frac{1-x^{2n}}{1-x^2}$

$\frac{1}{1-x^2} - 1 - x^2 - x^4 - ... - x^{2n-2}$

Then just integrate this which is straight forward.

Interesting method, nice work.

Alternatively, if we consider the integral to be I_n
We find that I(n-1)-I(n) = (something integratable)
Then take a sum of both sides from k=1 to n, the LHS telescopes to leave I(1) and I(n), the RHS being an unevaluatable sum.

braintic said:
I can only turn this into a recurrence relation leading to (ln3)/2 minus a sum of n terms, each term being 1/[(2k-1).2^k]. Is this correct, and is it possible to sum this series?

Yep that is the simplest form one can get it at HSC level, I think it might have something to do with harmonic numbers however if we were to evaluate it in terms of those terms.

Sy123 · Jul 7, 2013

Re: MX2 Integration Marathon

$\int \frac{x^2-1}{x\sqrt{x^4+1}} \ dx$

RealiseNothing · Jul 7, 2013

Re: MX2 Integration Marathon

Sy123 said:
$\int \frac{x^2-1}{x\sqrt{x^4+1}} \ dx$

This lol?

$\frac{1}{4}\ln(\frac{(x^4+2\sqrt{x^4+1}+2)(2x^4+2x^2+1)}{x^4}) + C$

RealiseNothing · Jul 7, 2013

Re: MX2 Integration Marathon

RealiseNothing said:
This lol?

$\frac{1}{4}\ln(\frac{(x^4+2\sqrt{x^4+1}+2)(2x^4+2x^2+1)}{x^4}) + C$

Keeping all the logs separate gives:

$\frac{1}{2}\ln(\sqrt{x^4+1} + x^2) + \frac{1}{4}\ln(\sqrt{x^4+1}+1) - \frac{1}{4}\ln(\sqrt{x^4+1}-1)$

Sy123 · Jul 7, 2013

Re: MX2 Integration Marathon

RealiseNothing said:
Keeping all the logs separate gives:

$\frac{1}{2}\ln(\sqrt{x^4+1} + x^2) + \frac{1}{4}\ln(\sqrt{x^4+1}+1) - \frac{1}{4}\ln(\sqrt{x^4+1}-1)$

I don't think that is correct, some of your logarithms look similar to my ones so you may have made a mistake somwhere.

asianese · Jul 7, 2013

Re: MX2 Integration Marathon

My answer was

$\dfrac{1}{2} \left( \ln\left[(x^2+\sqrt{x^4+1}\right] +\ln \left[ \dfrac{\sqrt{x^4+1}+1}{x^2}\right] \right) +\psi.$

Solution:

$$Let $ x^2=\tan \theta \Rightarrow x^2-1=\tan\theta -1, x^4=\tan^2 \theta, x^4+1=\sec^2 \theta.$

$2x \mathop{dx} = \sec^2 \theta \mathop{d\theta} \Rightarrow \mathop{dx} = \dfrac{\sec^2 \theta}{2\sqrt{\tan \theta}} \mathop{d\theta}.$

$\begin{align*}I&= \int \dfrac{\tan\theta -1}{\sqrt{\tan\theta}\sqrt{\sec^2 \theta}} \cdot \dfrac{\sec^2 \theta}{2\sqrt{\tan\theta}} \mathop{d\theta}\\ &= \int \dfrac{(\tan\theta -1)\sec\theta}{2\tan\theta} \mathop{d\theta} \\ &= \dfrac{1}{2} \int \dfrac{\tan\theta \sec\theta -\sec\theta}{\tan\theta} \mathop{d\theta} \\ &= \dfrac{1}{2} \int \sec \theta - \csc\theta \mathop{d\theta} \\ & =\dfrac{1}{2}\left[ \ln(\sec\theta + \tan\theta)+\ln(\csc\theta+\cot\theta)\right] \\ &=\dfrac{1}{2} \left[ \ln\left(\sqrt{x^4+1}+x^2\right)+\ln\left( \dfrac{\sqrt{x^4+1}+1}{x^2}\right) \right] + \psi. \end{align*}$

bleakarcher · Jul 7, 2013

Re: MX2 Integration Marathon

Sy123 said:
$\int \frac{x^2-1}{x\sqrt{x^4+1}} \ dx$

I got ln[(x+(1/x))+sqrt[x^2+(1/x)^2] + C

Sy123 · Jul 7, 2013

Re: MX2 Integration Marathon

bleakarcher said:
I got ln[(x+(1/x))+sqrt[x^2+(1/x)^2] + C

Yep this is what I was aiming at, divide both top and bottom by x^2, then substiute u=x+1/x, it dissolves into a standard integral.

Asianese and Realise your answers may be of a different form that can be manipulated using log properties or something.

bleakarcher · Jul 7, 2013

Re: MX2 Integration Marathon

Sy123 said:
Yep this is what I was aiming at, divide both top and bottom by x^2, then substiute u=x+1/x, it dissolves into a standard integral.

Asianese and Realise your answers may be of a different form that can be manipulated using log properties or something.

Thing is though, by dividing by x at the bottom and bottom then applying x=sqrt(x^2) we are assuming that x>0. So we've restricted the integrand to x>0 I think.

asianese · Jul 7, 2013

Re: MX2 Integration Marathon

I just checked in wolfram alpha - my answer was correct (differentiate the integral and get the same thing).

Scribbling some shit down with the logs you get the same answer as yours.

Sy123 · Jul 7, 2013

Re: MX2 Integration Marathon

bleakarcher said:
Thing is though, by dividing by x at the bottom and bottom then applying x=sqrt(x^2) we are assuming that x>0. So we've restricted the integrand to x>0 I think.

Yeah that's true....

Sy123 · Jul 7, 2013

Re: MX2 Integration Marathon

$Prove$

$\int_0^{\pi /2} \frac{dx}{1 + \cos \alpha \sin x} = \frac{\alpha}{\sin \alpha}$

$\cos \alpha \neq \pm 1$

asianese · Jul 7, 2013

Re: MX2 Integration Marathon

Sy123 said:
$Prove$

$\int_0^{\pi /2} \frac{dx}{1 + \cos \alpha \sin x} = \frac{\alpha}{\sin \alpha}$

Lel Syd Grammy 2011.

Sy123 · Jul 7, 2013

Re: MX2 Integration Marathon

asianese said:
Lel Syd Grammy 2011.

Originally STEP 200(something)!

Fun fact, some CSSA questions are directly ripped from a STEP paper. (i.e. 2011 Q7a, 2011 Q8b)
That is the only CSSA paper that I've seen so far, I'm going to start doing the other ones soon, but I'm guessing that even from the earlier CSSA papers there are STEP rips.

asianese · Jul 7, 2013

Re: MX2 Integration Marathon

Lel inception of maths questions.

braintic · Jul 7, 2013

Re: MX2 Integration Marathon

Sy123 said:
$Prove$

$\int_0^{\pi /2} \frac{dx}{1 + \cos \alpha \sin x} = \frac{\alpha}{\sin \alpha}$

Just checking - should there be a proviso that cos alpha is not equal to +/- 1 ?

HSC 2012-14 MX2 Integration Marathon (archive) (4 Viewers)

Well-Known Member

This too shall pass

Well-Known Member

what is that?It is Cowpea

This too shall pass

This too shall pass

what is that?It is Cowpea

what is that?It is Cowpea

This too shall pass

Σ

Active Member

This too shall pass

Active Member

Σ

This too shall pass

This too shall pass

Σ

This too shall pass

Σ

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 4)