Integration by reduction

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Davelittle
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#1
Report Thread starter 7 years ago
#1
 \int x^6 sinx\ dx between  0 and  \pi/2 .

So I carried out by parts twice and ended up with  6( \pi /2)^5 + 30  \int x^4 sinx\ dx meaning  I_n = 6(\pi/2)^{n-1}- 30I_{n-2}

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
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#2
(Original post by Davelittle)
 \int^0.5\pi_0(x^6 sinx \dx) between 0 and  \pi/2 .

So I carried out by parts twice and ended up with 6(pi/2)^5 + 30  \int(x^4 sinx dx) 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
post the question and some workings please
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Davelittle
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#3
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#3
(Original post by TeeEm)
post the question and some workings please
Let me sort out the latex one sec
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Phichi
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#4
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#4
(Original post by Davelittle)
 \int x^6 sinx \dx between  0 and  \pi/2 .

So I carried out by parts twice and ended up with  6( \pi /2)^5 + 30  \int x^4 sinx \dx 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
Its usually custom to do \int_0^{\frac{\pi}{2}} x^nsinx \,dx, then work out the case where n=6

Also, I've calculated the reduction formula to be;

I_n=n(\frac{\pi}{2})^{n-1}-n(n-1)I_{n-2}.

So from there we can see you're suffering a sign error. But to check I also went through the case where n=6 using parts, and ended up with:

I=6(\frac{\pi}{2})^5-30\int_0^{\frac{\pi}{2}} x^4sinx \,dx

So we can see, its definitely a sign error on your part.
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TeeEm
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#5
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#5
(Original post by Davelittle)
Let me sort out the latex one sec
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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Davelittle
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#6
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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
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#7
(Original post by Phichi)
Its usually custom to do \int_0^{\frac{\pi}{2}} x^nsinx \,dx, then work out the case where n=6

Also, I've calculated the reduction formula to be;

I_n=n(\frac{\pi}{2})^{n-1}-n(n-1)I_{n-2}.

So from there we can see you're suffering a sign error. But to check I also went through the case where n=6 using parts, and ended up with:

I=6(\frac{\pi}{2})^5-30\int_0^{\frac{\pi}{2}} x^4sinx \,dx

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.
Sorted it, thanks to you both thumbs up!
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Phichi
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#8
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#8
(Original post by Davelittle)
Sorted it, thanks to you both thumbs up!
No problem, just pay attention to those signs in by parts :P Also, although your method worked for starting with n=6, questions with reduction formulas will always expect you, or request you to find the reduction formula for I_n 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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#9
(Original post by Davelittle)
Sorted it, thanks to you both thumbs up!
goodnight
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