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

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Re: MX2 Integration Marathon

Are you sure?

My working got me down to



and so a can't be 0 since is undefined.
 

seanieg89

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Re: MX2 Integration Marathon

But in the initial integration, a=0 means that you are integrating over an interval of length 0, of course this will give you 0.
 

Sy123

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Re: MX2 Integration Marathon

the other solution is so painfullll
:p

Replace the roots with alpha and beta if you want, and anyway, I don't really want a solution I know people can do partial fractions.

The aim of the question was more getting people to recognize the different cases
 

JJ345

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Re: MX2 Integration Marathon

I derived the reduction formula using trig identities etc. But got a final answer of 2^(n/2) I(0)--> So what would my final answer be? I(0) is just a constant isn't it? Probably did something wrong.
 

Sy123

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Re: MX2 Integration Marathon

I derived the reduction formula using trig identities etc. But got a final answer of 2^(n/2) I(0)--> So what would my final answer be? I(0) is just a constant isn't it? Probably did something wrong.
The answer is:



This is because, (using sum to product formula)

So if n is even, it becomes I(0), and if n is odd, it becomes I(1)
 
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JJ345

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Re: MX2 Integration Marathon

The answer is:



This is because, (using sum to product formula)

So if n is even, it becomes I(0), and if n is odd, it becomes I(1)
Yeah, I didn't recognise the cases. But my final reduction formula was I(n)=2I(n-2) ???
 

Sy123

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Re: MX2 Integration Marathon

Yeah, I didn't recognise the cases. But my final reduction formula was I(n)=2I(n-2) ???
You may have made a silly mistake, because:






 

JJ345

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Re: MX2 Integration Marathon

You may have made a silly mistake, because:






Your right, but for some reason I still can't find the mistake in mine :/ Its wayyy longer than your method so probably did make a mistake somewhere along the line.
 
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