HSC 2012-14 MX2 Integration Marathon (archive) (5 Viewers)

Sy123 · May 11, 2013

Re: MX2 Integration Marathon

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

Yep, nice work.

asianese · May 11, 2013

Re: MX2 Integration Marathon

Geddit. $\psi\cdots$Sy$$ EHEHEHEHEHEHHHEEEHEH

Sy123 · May 11, 2013

Re: MX2 Integration Marathon

HeroicPandas said:
$$ Let e^x = cos2u, dx = $\frac{-2\sinu.du}{\cos{2u}} \\ \\ I = -2\int \sqrt{1+\cos2u}. \frac{\sin{u}}{\cos{2u}}du \\ I = -\sqrt{2} \int \tan(2u) du \\ \ I = \frac{\sqrt{2}}{2} ln(\cos2u) +C \\ I = \sqrt{2}ln(\sqrt{e^x}) + C$

What did i do wrong?

$dx \neq \frac{-2 \sin u}{\cos 2u} \ du$

asianese said:
Geddit. $\psi\cdots$Sy$$ EHEHEHEHEHEHHHEEEHEH

oo i geddit now, clever

HeroicPandas · May 11, 2013

Re: MX2 Integration Marathon

Sy123 said:
$dx \neq \frac{-2 \sin u}{\cos 2u} \ du$

oo i geddit now, clever

$e^x = \cos2u \\ $By implicit differentiation with respect to x, \\ \\ e^x .dx = -2\sin2u.du \\ \\dx = $\frac{-2sin2u}{e^x}du \\ $But e^x = cos2u \\ \\ \therefore dx = $\frac{-2sin2u}{cos2u}du$

So this is forbidden and incorrect?

Sy123 · May 11, 2013

Re: MX2 Integration Marathon

HeroicPandas said:
$e^x = \cos2u \\ $By implicit differentiation with respect to x, \\ \\ e^x .dx = -2\sin2u.du \\ \\dx = $\frac{-2sin2u}{e^x}du \\ $But e^x = cos2u \\ \\ \therefore dx = $\frac{-2sin2u}{cos2u}du$

So this is forbidden and incorrect?

Yes it is definitely correct, but note that what you used in your working out was sin(u) not sin(2u) like you have now.

HeroicPandas · May 11, 2013

Re: MX2 Integration Marathon

Sy123 said:
Yes it is definitely correct, but note that what you used in your working out was sin(u) not sin(2u) like you have now.

AHHH yes, thank u for spotting that

Registered User · May 11, 2013

Re: MX2 Integration Marathon

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

Is this correct?

Sy123 · May 11, 2013

Re: MX2 Integration Marathon

Registered User said:
Is this correct?

Maybe, sorry for being annoying but for this one I didn't do the question that gave out, I just look at it, determine what strategy should be used, deem if its possible etc etc, then post it. (only for integrals)

In this case, I would decide to do u=e^x, since du comes out easily. Then do partial fractions with the square root to split it up, then the substitution u = sin v, then from then on we have a trig integral, which might be computed with t=tanx/2 then partial fractions.
Or perhaps there is an easier way, I think I have stuffed up differentiating asin(e^x) - acos(e^x) + atan(e^x).

And Makematics already posted a solution on the previous page, maybe you should have a look at it?

Registered User · May 11, 2013

Re: MX2 Integration Marathon

Sy123 said:
Maybe, sorry for being annoying but for this one I didn't do the question that gave out, I just look at it, determine what strategy should be used, deem if its possible etc etc, then post it. (only for integrals)

In this case, I would decide to do u=e^x, since du comes out easily. Then do partial fractions with the square root to split it up, then the substitution u = sin v, then from then on we have a trig integral, which might be computed with t=tanx/2 then partial fractions.
Or perhaps there is an easier way, I think I have stuffed up differentiating asin(e^x) - acos(e^x) + atan(e^x).

And Makematics already posted a solution on the previous page, maybe you should have a look at it?

I did it using the trig substitution, but I'm just wondering if it can be done using the Abel transform $(u=\sqrt{X}')$

Makematics · May 12, 2013

Re: MX2 Integration Marathon

Sy123 said:
$\int_{- \infty}^{\infty} \frac{\ln(1+x^2)}{1+x^2} \ dx$

Guys I realised my answer to the question was wrong, i stuffed up a step and realised i cant get it. wanna give it a try? seems like a good question. i reduced it to a certain point but am now stuck :/

Registered User · May 12, 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:

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

Makematics · May 12, 2013

Re: MX2 Integration Marathon

Registered User said:

Ahhh right very nice

I got stuck on the 6th last line in my solution:L and still dont get it... How does the 1/2 change to a 1 and the limits change from pi to pi/2

ayecee · May 12, 2013

Re: MX2 Integration Marathon

Here's a nice one, It's part 2 that's giving me trouble. Anyone wana give it a shot?

Sy123 · May 12, 2013

Re: MX2 Integration Marathon

Registered User said:
http://i.imgur.com/JoNxwD2.png[img][/QUOTE]

Nice work.

[quote="ayecee, post: 6334624"]Here's a nice one, It's part 2 that's giving me trouble. Anyone wana give it a shot?
[ATTACH]28104[/ATTACH][/QUOTE]

[tex]$First denote from the previous part$ [/tex]

[tex]I_k = \int_0^{\pi/2} \cos x \sin^{2k}x \ dx = \frac{1}{2k+1} [/tex]

[tex]$ii)$ [/tex]

[tex]J = \int_{0}^{\pi/2} \cos x \cos^{2n}x \ dx = \int_0^{\pi /2} \cos x (1-\sin^2 x)^n \ dx [/tex]

[tex]$By the binomial theorem$ [/tex]

[tex]J = \int_{0}^{\pi /2} \cos x -\binom{n}{1} \cos x \sin^2 x + \binom{n}{2} \cos x \sin^4 x - \dots + (-1)^n \binom{n}{n} \cos x \sin^{2n}x \ dx [/tex]

[tex]J = I_0 - \binom{n}{1} I_1 + \binom{n}{2} I_n - \dots + (-1)^n \binom{n}{n} I_n [/tex]

[tex]J = \sum_{k=0}^{n} (-1)^k I_k \binom{n}{k} \equiv \sum_{k=0}^{n} \frac{(-1)^k}{2k+1} \binom{n}{k} [/tex]

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

[tex]$Prove that$ [/tex]

[tex]\int_0^1 \frac{x^4 (1-x)^4}{1+x^2} = \frac{22}{7} - \pi [/tex]

[tex]$Hence prove that$ \ \ \frac{22}{7} \ \ $exceeds$ \ \ \pi [/tex]

asianese · May 12, 2013

Re: MX2 Integration Marathon

What's even easier is make a substitution u=sinx and then you'll end up with bound of 0 and 1, with (1-u^2)^n which is easier to see where the binomials come in.

ayecee · May 12, 2013

Re: MX2 Integration Marathon

asianese said:
What's even easier is make a substitution u=sinx and then you'll end up with bound of 0 and 1, with (1-u^2)^n which is easier to see where the binomials come in.

That's what I did, then I didn't know what to do, havnt done binomial yet... I had a feeling it did due to the answer

Registered User · May 13, 2013

Re: MX2 Integration Marathon

This is the integration test taken by applicants to MIT undergrad courses. Have a go at it and post you solutions.

http://web.mit.edu/abhinavk/www/integrationbee/qual2012.pdf

Sy123 · May 13, 2013

Re: MX2 Integration Marathon

$\int_0^1 \ln \left(\sqrt{1-x} + \sqrt{1+x} \right ) \ dx$

HeroicPandas · May 13, 2013

Re: MX2 Integration Marathon

Sy123 said:
==========================

$Prove that$

$\int_0^1 \frac{x^4 (1-x)^4}{1+x^2} = \frac{22}{7} - \pi$

$$Hence prove that$ \ \ \frac{22}{7} \ \ $exceeds$ \ \ \pi$

did u get part of this from a sydney boys trial paper?

Sy123 · May 13, 2013

Re: MX2 Integration Marathon

HeroicPandas said:
did u get part of this from a sydney boys trial paper?

I just got it off a random site, as a proof that 22/7 exceeds pi.

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