1. ## Equivalencies when integrating?

So, often when integrating I end up using complex numbers, just because it seems obvious to me.
I don't get a wrong answer, just a different one, that marking criteria usually doesn't mention at all, so I'm just wondering if anyone else has this problem or if i should do something about it. I've attached an example just so you can see what I mean (sorry for bad quality)

21618219_751279525060196_1920454162_o.jpg

This is 2013 question 12 a) and it's worth a whopping 4 marks that I don't want to risk, if a marker disregards my answer just because it looks nothing like the 'correct' answer (which is ln(2)/3)
TIA

2. ## Re: Equivalencies when integrating?

I don't think you can use complex numbers. You can just split up the 1/(t^2 - 9) via inspection so it becomes:

(1/6) [1/(t-3) - 1/(t+3)] then you go from there using logs

EDIT: You might be right but in HSC pretty sure you can't do that

3. ## Re: Equivalencies when integrating?

Originally Posted by Skitzgerald
So, often when integrating I end up using complex numbers, just because it seems obvious to me.
I don't get a wrong answer, just a different one, that marking criteria usually doesn't mention at all, so I'm just wondering if anyone else has this problem or if i should do something about it. I've attached an example just so you can see what I mean (sorry for bad quality)

21618219_751279525060196_1920454162_o.jpg

This is 2013 question 12 a) and it's worth a whopping 4 marks that I don't want to risk, if a marker disregards my answer just because it looks nothing like the 'correct' answer (which is ln(2)/3)
TIA
Your answer is essentially still correct but you'll need to know about the complex arctangent function to simplify it down to the desired form of the answer (ln(2)/3). See for example here: http://scipp.ucsc.edu/~haber/archive...A10/arc_10.pdf . Knowledge of these matters is of course not within the HSC syllabus. To do the question in a "HSC" way, just use

$\int \frac{1}{t^2 -a^2}\,\mathrm{d}t = \frac{1}{2a}\ln \left|\frac{a-t}{a+t}\right| (with a = 3).$

You can derive this using partial fractions.

4. ## Re: Equivalencies when integrating?

Yea I know you can use partial fractions and that makes it easy, but I don't know why you wouldn't be able to use complex numbers, i's a constant and as long as you don't make an arithmetic error you're still gonna get a correct answer.
(2i/3)*tan^-1(-i/3) is equivalent to ln(2)/3 (according to wolfram alpha at least)
I understand that if the integral has a context, such as volume or distance then I obviously can't use complex numbers, but if its just an indefinite integral and my gut instinct is to force an inverse tan integral instead of using partial fractions then it'd be frustrating to lose marks because of it.

Ahh well, I guess I'll just try and avoid using complex numbers when integrating in future (unless of course I don't think i'll get an answer otherwise).
Thanks

5. ## Re: Equivalencies when integrating?

Originally Posted by Skitzgerald
Yea I know you can use partial fractions and that makes it easy, but I don't know why you wouldn't be able to use complex numbers, i's a constant and as long as you don't make an arithmetic error you're still gonna get a correct answer.
(2i/3)*tan^-1(-i/3) is equivalent to ln(2)/3 (according to wolfram alpha at least)
I understand that if the integral has a context, such as volume or distance then I obviously can't use complex numbers, but if its just an indefinite integral and my gut instinct is to force an inverse tan integral instead of using partial fractions then it'd be frustrating to lose marks because of it.

Ahh well, I guess I'll just try and avoid using complex numbers when integrating in future (unless of course I don't think i'll get an answer otherwise).
Thanks
Like I said, your answer involving complex numbers is essentially correct, but you need knowledge of the complex arctangent function to simplify it down to the desired form of the answer (ln(2)/3). And in the HSC, you should probably not use the complex method because it isn't in the syllabus and there's no real need to use it.

6. ## Re: Equivalencies when integrating?

Yea that was a reply to Pikachu.
Thanks for the link about the arctangent function, and i can understand that my answer isn't in its simplest form to anyone with knowledge about the arctan function, but everything i've done is mathematically correct and i've answered the question, so being confined to the "HSC" answer is annoying, i feel like im being punished for my creativity

7. ## Re: Equivalencies when integrating?

Originally Posted by Skitzgerald
Yea that was a reply to Pikachu.
Thanks for the link about the arctangent function, and i can understand that my answer isn't in its simplest form to anyone with knowledge about the arctan function, but everything i've done is mathematically correct and i've answered the question, so being confined to the "HSC" answer is annoying, i feel like im being punished for my creativity
Ah OK.

Well in any case, the markers are probably looking for the answer ln(2)/3. You should probably simplify your answer down to this because if you leave it as a complex arctangent, one could argue this isn't fully simplified, especially since we're dealing with a real integral (so we should give our final answer as a real number).

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