HSC 2012-14 MX2 Integration Marathon (archive) (1 Viewer)

Sy123 · May 11, 2013

Re: MX2 Integration Marathon

Registered User said:
From now on I will try to post only questions that are done by very tricky substitutions, IBP or very tedious.

$\int \frac{x^{2}}{(1+x^{2})^{2}}dx$

Upon the substitution:

$x= \tan \theta$

The integral resolves into:

$\int \sin^2 \theta \ d\theta$

$= \frac{1}{2} (\tan^{-1} x + \frac{x}{x^2+1}) + c$

====

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

asianese · May 11, 2013

Re: MX2 Integration Marathon

Registered User said:
$\int \cot^{-1}(x^{2}+x+1)dx$

Inverse cotangent isn't defined in the 4U Syllabus?

Registered User · May 11, 2013

Re: MX2 Integration Marathon

Sy123 said:
Upon the substitution:

$x= \tan \theta$

The integral resolves into:

$\int \sin^2 \theta \ d\theta$

$= \frac{1}{2} (\tan^{-1} x + \frac{x}{x^2+1}) + c$

Lol that was an easy one but I though it can be done by IBP be making u=x/2 etc. which is a bit harder

Registered User · May 11, 2013

Re: MX2 Integration Marathon

Sy123 said:
Yep, nice.

$\int_0^1 \frac{\ln(1+x^2)}{1+x^2} \ dx$

Awesome question!

$Now I need to evaluate$ $\frac{1}{2}\int_{0}^{\frac{\pi }{2}}ln(\sin u )du$

Sy123 · May 11, 2013

Re: MX2 Integration Marathon

Registered User said:
Awesome question!

$Now I need to evaluate$ $\frac{1}{2}\int_{0}^{\frac{\pi }{2}}ln(\sin u )du$

I don't think its a good integral man, its not evaluable in elementary functions, try this one instead:

$\int_{- \infty}^{\infty} \frac{\ln(1+x^2)}{1+x^2} \ dx$

Registered User · May 11, 2013

Re: MX2 Integration Marathon

Sy123 said:
I don't think its a good integral man, its not evaluable in elementary functions, try this one instead:

$\int_{- \infty}^{\infty} \frac{\ln(1+x^2)}{1+x^2} \ dx$

It can be done using elementary functions but it is a hard one... I will try the one with the new limits.

=========================

$\int_{0}^{1} \frac{\ln(1+x)}{1+x^2} \ dx$

Evaluate the above integral WITHOUT using the substitutions $x=tan\theta$ or $x=\frac{1-u}{1+u}$

Makematics · May 11, 2013

Re: MX2 Integration Marathon

Registered User said:
It can be done using elementary functions but it is a hard one... I will try the one with the new limits.

Sy's one is doable, i just needed several substitutions.

Sy123 · May 11, 2013

Re: MX2 Integration Marathon

Registered User said:
It can be done using elementary functions but it is a hard one... I will try the one with the new limits.

=========================

$\int_{0}^{1} \frac{\ln(1+x)}{1+x^2} \ dx$

Evaluate the above integral WITHOUT using the substitutions $x=tan\theta$ or $x=\frac{1-u}{1+u}$

The integral resolves into $-2\int_0^{\pi/4} \ln(\cos \theta) \ d\theta$

Which involves Catalan's Constant C

$C = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^2}$

And because of that I think it will involve Taylor series at some point due to the infinite sum.

Makematics · May 11, 2013

Re: MX2 Integration Marathon

Sy123 said:
I don't think its a good integral man, its not evaluable in elementary functions, try this one instead:

$\int_{- \infty}^{\infty} \frac{\ln(1+x^2)}{1+x^2} \ dx$

I get pi.ln2 -2
I dont have latex so i cant really post my sol but i can go through the procedure if you want.

EDIT: Just realised this is incorrect, fixing it up now.

Registered User · May 11, 2013

Re: MX2 Integration Marathon

Sy123 said:
The integral resolves into $-2\int_0^{\pi/4} \ln(\cos \theta) \ d\theta$

Which involves Catalan's Constant C

$C = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^2}$

And because of that I think it will involve Taylor series at some point due to the infinite sum.

True.

Registered User · May 11, 2013

Re: MX2 Integration Marathon

Sy123 said:
Upon the substitution:

$x= \tan \theta$

The integral resolves into:

$\int \sin^2 \theta \ d\theta$

$= \frac{1}{2} (\tan^{-1} x + \frac{x}{x^2+1}) + c$

====

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

I changed the integral to: $\int \frac{e^{x}}{(1+e^{2x})\sqrt{(1-e^{2x})}}dx$
The let u=e^x
The integral becomes: $\int \frac{du}{(1+u^{2})\sqrt{(1-u^{2})}}$
I can do this the long way, such as on wolfram alpha (let $u=\sin\theta$ ) but I think we can use a short cut to do it so let $u=\sqrt{1-u^{2}}'$

$\sqrt{1-u^{2}}'=-\frac{u}{\sqrt{1-u^2}} \therefore v^{2}=\frac{u^{2}}{1-u^{2}}$

$du=\frac{dv}{\sqrt{1-u^{2}}}$

The integral becomes: $\int \frac{dv}{1-u^{4}}$

I need to somehow get rid off the u and get the integral in terms of v so any idea on how this can be done?

Registered User · May 11, 2013

Re: MX2 Integration Marathon

Do you guys think it's a good idea to make a "Harder 3U marathon"? I think I have done enough integration, need to do some harder 3U.

Makematics · May 11, 2013

Re: MX2 Integration Marathon

Registered User said:
Do you guys think it's a good idea to make a "Harder 3U marathon"? I think I have done enough integration, need to do some harder 3U.

i dont see why not! we're approaching the pointy end of the year, and although most people wont do the topic until after their trials, i dont see any harm in starting now

Makematics · May 11, 2013

Re: MX2 Integration Marathon

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

Great question, but sorry for poor quality...

Final answer should read I = 1/sqrt2 arctan {e^x / sqrt [2(1- e^2x)]}

Registered User · May 11, 2013

Re: MX2 Integration Marathon

Makematics said:
View attachment 28098

Sorry for poor quality...

Do you think the shorter method is doable?

Sy123 · May 11, 2013

Re: MX2 Integration Marathon

I think its better that the Harder 3U questions are asked in the actual 4U marathon imo.

Most of the questions there are actually harder 3U if not polynomials

Makematics · May 11, 2013

Re: MX2 Integration Marathon

Registered User said:
Do you think the shorter method is doable?

not entirely sure, it looks doable, but i dont think it would be quicker. you'll have to confer with Sy.

Sy123 · May 11, 2013

Re: MX2 Integration Marathon

Makematics said:
View attachment 28098

Great question, but sorry for poor quality...

Final answer should read I = 1/sqrt2 arctan {e^x / sqrt [2(1- e^2x)]}

yep if it works, it works. To make that question that I actually differentiated:

$\frac{1}{2} (\sin^{-1}(e^x) - \cos^{-1}(e^x) + \tan^{-1}(e^x))$

So that may be another form for this.

Sy123 · May 11, 2013

Re: MX2 Integration Marathon

$\int \sqrt{1+e^{x}} \ dx$

asianese · May 11, 2013

Re: MX2 Integration Marathon

Sy123 said:
$\int \sqrt{1+e^{x}} \ dx$

$$Let $u^2=1+e^x \Longrightarrow e^x=u^2-1 \\ \ 2u\; du = e^x \; dx\\ \int \sqrt{1+e^x} \;dx \\ = \int u\cdot \frac{2u}{u^2-1} \; du\\ =2 \int \frac{u^2}{u^2-1} \; du \\ = 2\int 1+ \frac{1}{(u-1)(u+1)} \;du \\ = 2\int 1+\frac{1}{2(u-1)}-\frac{1}{2(u+1)} \; du\\ =2\left[u+\frac{1}{2} \ln ( u-1) -\frac{1}{2} \ln{(u+1)}\right] \\ =2\sqrt{1+e^x} +\ln (\sqrt{1+e^x}-1) - \ln (\sqrt{1+e^x}+1) + \bf{\psi}$

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