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

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Paradoxica

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Re: MX2 2015 Integration Marathon

Finally a question which isn't completely nuts
Although I do quite enjoy reading Ekman's and Paradoxica's solutions to some of their harder integrals, I admit.

Haha, I think "nuts" is an understatement :p

Anyway, for your question, change the denominator into the square of sine, split the integrand, and use IBP on the term containing the x. The new integral resulting from this cancels out the other term which we did not touch, leaving behind
 
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Ekman

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Re: MX2 2015 Integration Marathon

@porcupinetree







Edit: Paradoxica beat me to it :p
 

porcupinetree

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Re: MX2 2015 Integration Marathon

Here's a cool little one I came up with, probs too easy for y'all ;)

 

snail489

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Re: MX2 2015 Integration Marathon

Here's a cool little one I came up with, probs too easy for y'all ;)

Split an x^2 off the numerator into two fractions, then use reverse chain rule on the first while the second one factors into a perfect square on the denominator to get a ln expression. Should be something like (2/3)sqrt(x^6+14x^3+66) + (1/3)ln(x^3+7+sqrt(x^6+14x^3+66))
 

Carrotsticks

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Re: MX2 2015 Integration Marathon

A Combination of Intuition, Experience, Hindsight and Luck. Symbolic Integration is an Art, not a Science. Numerical Integration is a Science. Specifically a Computer Science. Okay I'll stop Misusing Definitions now.
Could you also stop misusing capitals whilst you're at it? =)
 

snail489

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Re: MX2 2015 Integration Marathon

Urgh that was horrible. Maybe I did it in some inefficient way but when I went to verify my answer with wolfram.... it gave me this gigantic expression with complex numbers...
Here's what I got:
82340e6818.png

I did this by expanding the sin(4x) thing and factorising the denominator by the difference of two squares, then using partial fractions with the sin and cos stuff (lol) and then using a t-formula substitution afterwards to finally integrate it... yeah that doesn't sound efficient at all.
 

Carrotsticks

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Re: MX2 2015 Integration Marathon

Haha you guys.

Urgh that was horrible. Maybe I did it in some inefficient way but when I went to verify my answer with wolfram.... it gave me this gigantic expression with complex numbers...
Here's what I got:
View attachment 32473

I did this by expanding the sin(4x) thing and factorising the denominator by the difference of two squares, then using partial fractions with the sin and cos stuff (lol) and then using a t-formula substitution afterwards to finally integrate it... yeah that doesn't sound efficient at all.
Similar idea to what I did, though I'm not too sure why you needed t formula substitutions at all.

Use double angle formulas to cancel as much as you can to get 1/cos(x)cos(2x). Multiply top/bottom by cos, turn the denominator into sines, and that is just a standard partial fractions problem (if you like, you can let u=sin(x) to see it more clearly).
 

Paradoxica

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Re: MX2 2015 Integration Marathon

Haha you guys.



Similar idea to what I did, though I'm not too sure why you needed t formula substitutions at all.

Use double angle formulas to cancel as much as you can to get 1/cos(x)cos(2x). Multiply top/bottom by cos, turn the denominator into sines, and that is just a standard partial fractions problem (if you like, you can let u=sin(x) to see it more clearly).
Now that's what I was going for!
 

Ekman

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Re: MX2 2015 Integration Marathon

Hey guys can someone solve this integral?



For some reason the answers for the paper for this question have +-, can someone explain why?
 

Paradoxica

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Re: MX2 2015 Integration Marathon

Sorry Ekman, the LaTeX isn't working right now (once again). Actually, has anyone ever contacted codecogs to see if this is an issue on their side? Anyway Ekman, can you send an image form of your question?
 

psyc1011

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Re: MX2 2015 Integration Marathon

Hey guys can someone solve this integral?



For some reason the answers for the paper for this question have +-, can someone explain why?
Was it +-sec^-1 x + C?
 

Drsoccerball

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Re: MX2 2015 Integration Marathon

Hey guys can someone solve this integral?



For some reason the answers for the paper for this question have +-, can someone explain why?
Did they say the answer was : tan^{-1}(+- \sqrt{x^2-1}) +C
 

psyc1011

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Re: MX2 2015 Integration Marathon

Precisely this yes.
The substitution (x = sec(u)) morphs the integrands to ∫tanu/(|tanu|) du and the sign is dependent on the values of x or bounds of the original integral

Either I've overlooked something or you overlooked something lol
 
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