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HSC 2012-14 MX2 Integration Marathon (archive) (1 Viewer)

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RealiseNothing

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Re: MX2 Integration Marathon

I think something might be wrong, i got some weird shit with the limits too :/
The integral doesn't converge lol, although if you were to do an indefinite integral I'm fairly sure what I got is right.
 

RealiseNothing

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Re: MX2 Integration Marathon

Posted this in 3U thread but why not:

Show that:

 
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Sy123

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Re: MX2 Integration Marathon

Using the substitution:



Resolves the integral to:



Upon the subtitution:



The integral resolves to:



Upon computation:



Substituting u back in:



Substituting back in x:

 
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Re: MX2 Integration Marathon

Alternatively you could have rationalised the numerator and put x^2=sin@.
 

Trebla

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Re: MX2 Integration Marathon



Where n is some positive integer
 

Sy123

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Re: MX2 Integration Marathon



Where n is some positive integer












Is this the answer you expected or is it possible to find that sum? Because I guess you could find that sum in integral form by integrating a geometric series of common ratio sin^2 x
But I think that just leads to the same thing as when you integrate by parts.
 

Trebla

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Re: MX2 Integration Marathon













Is this the answer you expected or is it possible to find that sum? Because I guess you could find that sum in integral form by integrating a geometric series of common ratio sin^2 x
But I think that just leads to the same thing as when you integrate by parts.
Pretty much the answer I was after. I also had in mind a bit of manipulation to generate the series as an alternative approach.

 

jmk123

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Re: MX2 Integration Marathon

I don't think so, what was your method briefly?
let u^3=x, 3u^2du=dx

So integral is 3u^2/(u^3/2+u).du
take out factor of u on bottom, it cancels with one u on top so u have

3u/(sqrtu+1)

let x=sqrtu and divide 3x^2 by (x+1), get -3/3x^2 as remainder, so -1/x^2, and the rest expands to 3x^2-3

So just integrated each bit separately.
 

Sy123

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Re: MX2 Integration Marathon

let u^3=x, 3u^2du=dx

So integral is 3u^2/(u^3/2+u).du
take out factor of u on bottom, it cancels with one u on top so u have

3u/(sqrtu+1)

let x=sqrtu and divide 3x^2 by (x+1), get -3/3x^2 as remainder, so -1/x^2, and the rest expands to 3x^2-3

So just integrated each bit separately.
Upon your substitution, , you didn't find du.
 

jmk123

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Re: MX2 Integration Marathon

Upon your substitution, , you didn't find du.
K I did that and got int(6x^3+6-(1/((x-0.5)^2+0.75)))


Final result is (3x^4)/2+6x-(2/sqrt3).tan^-1((2x-1)/sqrt3)
That right?
 
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