HSC 2015 MX2 Integration Marathon (archive) (1 Viewer)

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Drsoccerball

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Re: MX2 2015 Integration Marathon

So you literally have no choice but to stare at the question until you see reverse quotient rule?
...
Wow
...
If there were bounds we may have been able to do something else but since it doesnt i wouldn't see any other way...
 

InteGrand

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Re: MX2 2015 Integration Marathon

So you literally have no choice but to stare at the question until you see reverse quotient rule?
...
Wow
...
Whenever you see something squared on the denominator of the integrand, you should be thinking of the reverse quotient rule (especially if it's an indefinite integral).
 

leehuan

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Re: MX2 2015 Integration Marathon

Whenever you see something squared on the denominator of the integrand, you should be thinking of the reverse quotient rule (especially if it's an indefinite integral).
That wasn't the hard part. The hard part was realising that the product rule got sandwiched in like that, and that a +sinxcosx-sinxcosx was needed. The Pythagorean identity was not bad I suppose.
 

kawaiipotato

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Re: MX2 2015 Integration Marathon






 
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braintic

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Re: MX2 2015 Integration Marathon

Experimenting with latex - so this could be rubbish:

 

braintic

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Re: MX2 2015 Integration Marathon

Was in the process, but full working:


When doing that recurrence relation, it is a little bit slicker if you start with I_n + I_(n-2).
 

leehuan

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Re: MX2 2015 Integration Marathon

When doing that recurrence relation, it is a little bit slicker if you start with I_n + I_(n-2).
Touche, forgot about that
_______________________






Got too lazy from there. Clearly A = 1/sqrt(6) though.
 

Paradoxica

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Re: MX2 2015 Integration Marathon

Wolfram Integrator confirms 1/√6
yep, i did it on paper and got the correct answer.
hint: differentiation is much faster than integration for that problem.
Next problem:
evaluate:

Wolfram Alpha gives the answer in terms of elliptic integrals, but from my sources, an elementary solution definitely exists
 
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InteGrand

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Re: MX2 2015 Integration Marathon

yep, i did it on paper and got the correct answer.
hint: differentiation is much faster than integration for that problem.
Next problem:
evaluate:

Wolfram Alpha gives the answer in terms of elliptic integrals, but from my sources, it can be done elementarily (is that even a word?).
If you have an apparently elementary solution, just differentiate it to check if it works.
 

Drsoccerball

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Re: MX2 2015 Integration Marathon

I got a way to do it but it takes waaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaay tooo long
 

Drsoccerball

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Re: MX2 2015 Integration Marathon

If i can integrate id have the answer... Do you use partial fractions (x^2-2^(1/2)x+1)(x^2+2^(1/2)x +1)
 

Paradoxica

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Re: MX2 2015 Integration Marathon

If i can integrate id have the answer... Do you use partial fractions (x^2-2^(1/2)x+1)(x^2+2^(1/2)x +1)
yeah, that's a standard rational integral which can be decomposed into partial fractions. If I gave you the answer, do you think you might be able to spot a trick by differentiating it?
 

Drsoccerball

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Re: MX2 2015 Integration Marathon

yeah, that's a standard rational integral which can be decomposed into partial fractions. If I gave you the answer, do you think you might be able to spot a trick by differentiating it?
Hmmm maybe but for now heres my working out :









 
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nightweaver066

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Re: MX2 2015 Integration Marathon

yep, i did it on paper and got the correct answer.
hint: differentiation is much faster than integration for that problem.
Next problem:
evaluate:

Wolfram Alpha gives the answer in terms of elliptic integrals, but from my sources, an elementary solution definitely exists
Pretty ugly. You can use and for each integral respectively.

Simplify etc. etc. Then use to end up with something like which we can quickly decompose (using our best mate the Heaviside Cover-Up Method), integrate, then get rid of our substitutions.
 
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