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Student 999
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#1
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#1
struggling to show the (1+t^-2)/2 part, I've tried showing it directly and also tried showing it from the RHS but can't seem to factorise into anything usefulName:  Screenshot 2022-06-25 at 16.49.11.png
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RandomE
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#2
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#2
(Original post by Student 999)
struggling to show the (1+t^-2)/2 part, I've tried showing it directly and also tried showing it from the RHS but can't seem to factorise into anything usefulName:  Screenshot 2022-06-25 at 16.49.11.png
Views: 16
Size:  74.6 KBName:  IMG_8E8FF96D204C-1.jpeg
Views: 18
Size:  98.9 KB
I’d say let the x^2+1=t instead of the full thing, if you want me to follow that through to see if i get anywhere let me know.
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Student 999
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#3
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#3
(Original post by RandomE)
I’d say let the x^2+1=t instead of the full thing, if you want me to follow that through to see if i get anywhere let me know.
Don't think that'll work since you want the f(t) group aswell
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Student 999
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I've managed to show it now but would there have been a shorter way about it?
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Student 999
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#5
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#5
For the second part I could directly just do my substitution from part a and get the answer, what I did is referenced part a to save time but my explanation on how they're linked feels mathematically weak?Name:  IMG_A470F2C20473-1.jpeg
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RandomE
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#6
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#6
Don’t reckon so
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mqb2766
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#7
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(Original post by Student 999)
For the second part I could directly just do my substitution from part a and get the answer, what I did is referenced part a to save time but my explanation on how they're linked feels mathematically weak?Name:  IMG_A470F2C20473-1.jpeg
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Id guess that would be the solution they're expecting. Not sure why you think its weak.
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Student 999
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#8
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#8
(Original post by mqb2766)
Id guess that would be the solution they're expecting. Not sure why you think its weak.
I meant as in the why I can directly use f(t)= t^-3, would I not need to give some form of explanation?
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mqb2766
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#9
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#9
(Original post by Student 999)
I meant as in the why I can directly use f(t)= t^-3, would I not need to give some form of explanation?
Thats the form of the function in the original integral, so youre simply applying the general result from part a) to a particular function.
Last edited by mqb2766; 1 month ago
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