HSC 2012-14 MX2 Integration Marathon (archive) (2 Viewers)

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Carrotsticks

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Re: MX2 Integration Marathon

Just checking - should there be a proviso that cos alpha is not equal to +/- 1 ?
Maybe a range of alpha such as 0 < alpha < pi, which I can imagine would be a more 'standard condition' than cos alpha =/= plusminus 1.
 

Makematics

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Re: MX2 Integration Marathon

lol to be fair sydney grammar leads you through it... and that is a q8 :O 7 marks for it too!
 

Sy123

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Re: MX2 Integration Marathon

lol to be fair sydney grammar leads you through it... and that is a q8 :O 7 marks for it too!
'newer papers are easier' ; )

Seriously though, the tan(x/2) substitution is obvious, and the tan inverse identity is obvious, the only thing even requiring a sort of intuition is the use of double angle formulae while simplifying which in a sense is obvious when we observe what we are supposed to get, we can predict the steps needed to arrive at it!

7 marks is a bit much though..... and to think Sydney Grammar papers are usually the best out of top selective school papers.
 
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Re: MX2 Integration Marathon

TBH Syd grammar papers are one of the best out there...not just because they have pretty diagrams and are written in TeX...
 

Sy123

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Re: MX2 Integration Marathon

TBH Syd grammar papers are one of the best out there...not just because they have pretty diagrams and are written in TeX...
Yeah, relative to most other schools, definitely.
 

Carrotsticks

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Re: MX2 Integration Marathon

TBH Syd grammar papers are one of the best out there...not just because they have pretty diagrams and are written in TeX...
2012 was a bit of a disappoint though, I must say.
 
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Re: MX2 Integration Marathon

An easier question: I guess leave for someone else (nonregular..?)





 
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Sy123

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Re: MX2 Integration Marathon

Oops forgot limits.

Also just had a look at 2012 Syd Grammar. It resembled a random school paper. What happened?....Gosh.
 

Sy123

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Re: MX2 Integration Marathon

Well done
The idea is correct, but there is a mistake when you substitute cos(k) back in, it is not:



But rather the square root of that expression.

An easier substitution is:



Then re-arranging:



It then becomes a partial fraction problem.
 

Sy123

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Re: MX2 Integration Marathon









 

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Re: MX2 Integration Marathon

Well done
The idea is correct, but there is a mistake when you substitute cos(k) back in, it is not:



But rather the square root of that expression.
I think you misread my answer sir.
 

Sy123

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Re: MX2 Integration Marathon

1. Use the substitution u^2=x and so 2u du = dx

2. Substitute into the integrand.

2. Let v^2 = b+u (so that u=v^2 - b) and 2v dv = du.

3. Substitute into the integrand.

4. Let w^2 = a+v (so that v=w^2 -a) and 2w dw = dv.

5. Substitute into the integrand.

6. Expand all the factors in terms of w.

7. Integrate wrt w.

8. Do a mass back substitution.

9. Add a constant.






 
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Sy123

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Re: MX2 Integration Marathon

Using the identity:





Therefore:



Now, upon the generic substitution:

, we arrive at:



The integral will always be in the form, P/Q for some polynomials P and Q since the only operations done is the division and multiplication of polynomials with polynomials, and hence the end result will be in the form P/Q for some polynomials.

Therefore by the first theorem given in the question, the integral is evaluatable.

(I do realise that there is a possibility that the integral might not converge, I apologise for this inconsistency).


1. Use the substitution u^2=x and so 2u du = dx

2. Substitute into the integrand.

2. Let v^2 = b+u (so that u=v^2 - b) and 2v dv = du.

3. Substitute into the integrand.

4. Let w^2 = a+v (so that v=w^2 -a) and 2w dw = dv.

5. Substitute into the integrand.

6. Expand all the factors in terms of w.

7. Integrate wrt w.

8. Do a mass back substitution.

9. Add a constant.






Yep, alternatively you can do the substitution:



Which essentially combines all the substitutions you have used into a nice clean one.

For your integral, upon the substitution: , dissolves into the integral of cosecant of u
We can now do the t=tan u/2 substitution or just recognise it to be a standard integral not standard to the HSC, resulting in:



=====

 
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