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Integration by reduction
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Davelittle
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#1
between 
and 
.So I carried out by parts twice and ended up with
meaning 
Then I carried this out this reduction formulae to my integral and didn't end up with the same answer as wolfram alpha.
Can someone check what I've done wrong please
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TeeEm
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#2
(Original post by Davelittle)
between 0 and .
So I carried out by parts twice and ended up with 6(pi/2)^5 + 30
meaning In= (pi/2)^n-1 - 30In-2 (the In is capital I subscript n not natural log).
Then I carried this out this reduction formulae to my integral and didn't end up with the same answer as wolfram alpha.
Can someone check what I've done wrong please
between 0 and .So I carried out by parts twice and ended up with 6(pi/2)^5 + 30
meaning In= (pi/2)^n-1 - 30In-2 (the In is capital I subscript n not natural log).Then I carried this out this reduction formulae to my integral and didn't end up with the same answer as wolfram alpha.
Can someone check what I've done wrong please
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Davelittle
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#3
(Original post by TeeEm)
post the question and some workings please
post the question and some workings please
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Phichi
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#4
(Original post by Davelittle)
between and .
So I carried out by parts twice and ended up with 
meaning In= (pi/2)^n-1 - 30In-2 (the In is capital I subscript n not natural log).
Then I carried this out this reduction formulae to my integral and didn't end up with the same answer as wolfram alpha.
Can someone check what I've done wrong please
between and .So I carried out by parts twice and ended up with
meaning In= (pi/2)^n-1 - 30In-2 (the In is capital I subscript n not natural log).Then I carried this out this reduction formulae to my integral and didn't end up with the same answer as wolfram alpha.
Can someone check what I've done wrong please
, then work out the case where 
Also, I've calculated the reduction formula to be;
.So from there we can see you're suffering a sign error. But to check I also went through the case where
using parts, and ended up with:
So we can see, its definitely a sign error on your part.
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TeeEm
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#5
(Original post by Davelittle)
Let me sort out the latex one sec
Let me sort out the latex one sec
look at questions 11 and 19 in this link
http://madasmaths.com/archive/maths_...n_formulas.pdf
they are very similar.
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Davelittle
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#6
I realise that starting with I6 is incorrect, got it now thanks for the quick replies!
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Davelittle
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#7
(Original post by Phichi)
Its usually custom to do, then work out the case where
Also, I've calculated the reduction formula to be;
.
So from there we can see you're suffering a sign error. But to check I also went through the case where using parts, and ended up with:
So we can see, its definitely a sign error on your part.
Its usually custom to do
, then work out the case where Also, I've calculated the reduction formula to be;
.So from there we can see you're suffering a sign error. But to check I also went through the case where
using parts, and ended up with:So we can see, its definitely a sign error on your part.
(Original post by TeeEm)
I will be off very soon.
look at questions 11 and 19 in this link
http://madasmaths.com/archive/maths_...n_formulas.pdf
they are very similar.
I will be off very soon.
look at questions 11 and 19 in this link
http://madasmaths.com/archive/maths_...n_formulas.pdf
they are very similar.
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Phichi
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#8
(Original post by Davelittle)
Sorted it, thanks to you both thumbs up!
Sorted it, thanks to you both thumbs up!
, questions with reduction formulas will always expect you, or request you to find the reduction formula for 
to begin with. Secondly, some reductions you'll be unable to find by beginning with a certain n value, and they require a lot of manipulation.
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TeeEm
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#9
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