juantheron
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Re: MX2 2015 Integration Marathon

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So if you rearrange you get :
Can you really make that substitution when clearly cot 0 is undefined?So if you rearrange you get :
Let u^2=cotx and after that your integral becomes:
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Can you really make that substitution when clearly cot 0 is undefined?
Also, can you show ur working for how you went from the trig to the u bit? I don't think ur integrand is quite right....
That's exactly what i got, except with a negative sign
I had this as well. But the partial fraction decomposition for this is ridiculously tedious.
That's exactly what i got, except with a negative sign
So if you rearrange you get :
Let u^2=cotx and after that your integral becomes:
![]()
![]()
The algebra mistakes are real
Hm..
The algebra mistakes are real
This was beautifulEvaluate
"" is an arbitrary constant
Evaluate
"" is an arbitrary constant
the answersLol it's positive everywhere, its integral over the real line is definitely NOT 0.
That doesn't converge either.I did it from![]()
Doesnt it give you 0 ? And also wot u mean by converging and diverging etc..That doesn't converge either.