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Integration by reduction

∫x6sinx dx \int x^6 sinx\ dx between 0 0 and π/2 \pi/2 .

So I carried out by parts twice and ended up with 6(π/2)5+30 6( \pi /2)^5 + 30 ∫x4sinx dx \int x^4 sinx\ dx meaning In=6(π/2)n−1−30In−2 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
(edited 9 years ago)
Reply 1
Avatar for TeeEm
TeeEm
19
Original post by Davelittle
Unparseable latex formula:

\int^0.5\pi_0(x^6 sinx \dx)

between 0 and π/2 \pi/2 .

So I carried out by parts twice and ended up with 6(pi/2)^5 + 30 ∫(x4sinxdx) \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
Reply 2
Avatar for Davelittle
Davelittle
OP
13
Original post by TeeEm
post the question and some workings please


Let me sort out the latex one sec
Reply 3
Avatar for Phichi
Phichi
11
Original post by Davelittle
Unparseable latex formula:

\int x^6 sinx \dx

between 0 0 and π/2 \pi/2 .

So I carried out by parts twice and ended up with 6(π/2)5+30 6( \pi /2)^5 + 30
Unparseable latex formula:

\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 ∫0π2xnsinx dx\int_0^{\frac{\pi}{2}} x^nsinx \,dx, then work out the case where n=6n=6

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

In=n(π2)n−1−n(n−1)In−2I_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=6n=6 using parts, and ended up with:

I=6(π2)5−30∫0π2x4sinx dxI=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.
(edited 9 years ago)
Reply 4
Avatar for TeeEm
TeeEm
19
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_booklets/further_topics/integration/reduction_formulas.pdf

they are very similar.
Reply 5
Avatar for Davelittle
Davelittle
OP
13
I realise that starting with I6 is incorrect, got it now thanks for the quick replies!
Reply 6
Avatar for Davelittle
Davelittle
OP
13
Original post by Phichi
Its usually custom to do ∫0π2xnsinx dx\int_0^{\frac{\pi}{2}} x^nsinx \,dx, then work out the case where n=6n=6

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

In=n(π2)n−1−n(n−1)In−2I_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=6n=6 using parts, and ended up with:

I=6(π2)5−30∫0π2x4sinx dxI=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_booklets/further_topics/integration/reduction_formulas.pdf

they are very similar.


Sorted it, thanks to you both thumbs up!
Reply 7
Avatar for Phichi
Phichi
11
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=6n=6, questions with reduction formulas will always expect you, or request you to find the reduction formula for InI_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.
Reply 8
Avatar for TeeEm
TeeEm
19
Original post by Davelittle
Sorted it, thanks to you both thumbs up!


goodnight

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