HSC 2016 MX1 Marathon (archive) (3 Viewers)

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Paradoxica

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Re: HSC 2016 3U Marathon

Trigonometric substitutions were already split off from simplistic substitutions in 4U/3U for the difficulty of having to sub back in. So I'm just going to spoil some of the answer.




And at this point I'm still not sure how this is done using only 3U methods
Do not consider the substitution right away. Split the integral apart by rationalising the denominator before attempting anything else, as you will obtain two fairly simple rational terms. Then proceed with the substitution.
 

leehuan

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Re: HSC 2016 3U Marathon

Do not consider the substitution right away. Split the integral apart by rationalising the denominator before attempting anything else, as you will obtain two fairly simple rational terms. Then proceed with the substitution.
For some reason when I processed the rationalising method in my head it didn't yield something that was more simplified than the above result. May have to give it an actual attempt later.
 

leehuan

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Re: HSC 2016 3U Marathon

That csc^2(theta)sec(theta) was the most problematic one I observed for MX1. That's why I didn't seek to continue
 

Paradoxica

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Re: HSC 2016 3U Marathon

That csc^2(theta)sec(theta) was the most problematic one I observed for MX1. That's why I didn't seek to continue
Well from what KoA has told me, the way he expects an X1 student to do it is through advanced algebraic manipulation.

Basically partial fractions by inspection.
 

KingOfActing

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Re: HSC 2016 3U Marathon

I'm at school right now so I can't type it up, but the way to solve it is to NOT split the integral at all.

Keep it as is, until you end up with (sinθ +1)cosθ/((1-sinθ)sin^3(θ))

Now let u = sinθ

Next is adding and subtracting u or 2u for "partial fraction"ish part
 

leehuan

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Re: HSC 2016 3U Marathon

I'm at school right now so I can't type it up, but the way to solve it is to NOT split the integral at all.

Keep it as is, until you end up with (sinθ +1)cosθ/((1-sinθ)sin^3(θ))

Now let u = sinθ

Next is adding and subtracting u or 2u for "partial fraction"ish part

This is 3U. You have to give ALL the substitutions.


But yep I didn't think far enough to consider all the algebraic manipulations
 

Blitz_N7

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Re: HSC 2016 3U Marathon

I don't think any of those exist. ;)

BY ONLY SUBSTITUTION:
(1/2)lnsecarctan(sqrt(x-2)/2)+lnsinarctan(sqrt(x-2)/2)-1/4csc^2arctan(sqrt(x-2)/2)-(1/2)lnsinarctan(sqrt(x-2)/2)+ln(secarctan(sqrt(x-2)/2)+tanarctan(sqrt(x-2)/2))+1/2lntanarctan(sqrt(x-2)/2)-cscarctan(sqrt(x-2)/2)
 

leehuan

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Re: HSC 2016 3U Marathon

BY ONLY SUBSTITUTION:
(1/2)lnsecarctan(sqrt(x-2)/2)+lnsinarctan(sqrt(x-2)/2)-1/4csc^2arctan(sqrt(x-2)/2)-(1/2)lnsinarctan(sqrt(x-2)/2)+ln(secarctan(sqrt(x-2)/2)+tanarctan(sqrt(x-2)/2))+1/2lntanarctan(sqrt(x-2)/2)-cscarctan(sqrt(x-2)/2)
I'm sorry, but this just looked like gibberish.
 

si2136

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Re: HSC 2016 3U Marathon

Does anyone know how to graph Rational Functions without drawing a table of values, and just by analysing the equation?
 
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