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

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leehuan

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Re: MX2 2015 Integration Marathon



Sure. I wouldn't have made it past there though. Periodicity of trigonometric functions in integrals meant I would've had to draw a graph first.
 

Ekman

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Re: MX2 2015 Integration Marathon

I was able to manipulate the question down to this integral:



If someone can find a way to make it equal to then the integral can be solved
 

Carrotsticks

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Re: MX2 2015 Integration Marathon



My bank of integral tricks isn't really that large, but if I try adding those two integrals up I don't get anywhere either..

And wolfram does give an answer. pi log2.
Continuing on from what you have, that integral becomes



And this integral (ignoring the -2 at the front) evaluates to , which gives the answer. But the only way I know of evaluating this is the classic method taught in complex analysis (analytically continuing the integral of a power of sine and differentiating the resultant gamma expression). There may be a highschool friendly way of doing this, but none that I know of at the moment.
 

Drsoccerball

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Re: MX2 2015 Integration Marathon

Continuing on from what you have, that integral becomes



And this integral (ignoring the -2 at the front) evaluates to , which gives the answer. But the only way I know of evaluating this is the classic method taught in complex analysis (analytically continuing the integral of a power of sine and differentiating the resultant gamma expression). There may be a highschool friendly way of doing this, but none that I know of at the moment.
Someone solved using elementary methods on one of the previous pages.
 

seanieg89

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Re: MX2 2015 Integration Marathon

 
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kawaiipotato

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Re: MX2 2015 Integration Marathon

Continuing on from what you have, that integral becomes



And this integral (ignoring the -2 at the front) evaluates to , which gives the answer. But the only way I know of evaluating this is the classic method taught in complex analysis (analytically continuing the integral of a power of sine and differentiating the resultant gamma expression). There may be a highschool friendly way of doing this, but none that I know of at the moment.
I've seen people do it by using int f(pi/2 - x) = int f(x)
giving I = int 0 to pi/2 of ln(costheta) dtheta
Then adding, giving 2I = int 0 to pi/2 of ln(sin2theta) - ln2 dtheta
Then using u = 2theta, 2I = int 0 to pi of ln(sinu) - ln2 du
And then they evaluate the int 0 to pi of ln(sinu) to be zero because it's an odd around theta = pi/2 or something
 

seanieg89

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Re: MX2 2015 Integration Marathon

I've seen people do it by using int f(pi/2 - x) = int f(x)
giving I = int 0 to pi/2 of ln(costheta) dtheta
Then adding, giving 2I = int 0 to pi/2 of ln(sin2theta) - ln2 dtheta
Then using u = 2theta, 2I = int 0 to pi of ln(sinu) - ln2 du
And then they evaluate the int 0 to pi of ln(sinu) to be zero because it's an odd around theta = pi/2 or something
You are mostly right, not the last line though. Will fix up my eqn which isn't displaying for some reason.
 

Carrotsticks

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Re: MX2 2015 Integration Marathon

Someone solved using elementary methods on one of the previous pages.
Actually I think I know what you are referring to. I remember seeing something like that a while ago, and iirc braintic made a pdf about it too.
 

Drsoccerball

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Re: MX2 2015 Integration Marathon

Actually I think I know what you are referring to. I remember seeing something like that a while ago, and iirc braintic made a pdf about it too.
Sean did it just up there^
 

seanieg89

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Re: MX2 2015 Integration Marathon

Why is I = int 0 to pi/2 ln(sinx)dx = int 0 to pi/2 ln(sin2x) dx ?


You should of course be painstakingly slow in the jumps in your working for your HSC exams, I am just lazy / students trying to fill in blanks is actually decent exercise.
 
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Drsoccerball

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Re: MX2 2015 Integration Marathon



You should of course be painstakingly slow in the jumps in your working for your HSC exams, I am just lazy / students trying to fill in blanks is actually decent exercise.
Instead of (a-x) couldn't you just say that the area under sinx from 0 to pi/2 is the same as pi/2 to pi ?
 

kawaiipotato

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Re: MX2 2015 Integration Marathon

Instead of (a-x) couldn't you just say that the area under sinx from 0 to pi/2 is the same as pi/2 to pi ?
How wouldyou prove that it's the same besides using substitution: u = pi/2 - x?
 

seanieg89

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Re: MX2 2015 Integration Marathon

Instead of (a-x) couldn't you just say that the area under sinx from 0 to pi/2 is the same as pi/2 to pi ?
But why is the area under sin(x) between 0 and pi/2 the same as that between pi/2 and pi?

This transformation is exactly how you would prove this rigorously (*).


(*) At least as rigorously as possible in high school. A more subtle point here is that area is not really rigorously defined in high school. Before getting to more advanced notions of volume in measure theory, we typically define the area only for certain regions. (Such as those bounded by the x axis, two vertical lines, and the graph of a positive continuous function f(x). For such regions we would literally DEFINE area as being the definite integral of this function between the two vertical lines.)
 
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