intergration by parts

i have done a few intergration by parts questions and I am struggling to find why some equations need to have the process done twice

any explanation would be appreciated.

Reply 1

davros

Original post by mikeabc

i have done a few intergration by parts questions and I am struggling to find why some equations need to have the process done twice

any explanation would be appreciated.

Hard to explain without a specific example, but note that:
(i) there's no guarantee that IBP will work at all in the general case!
(ii) if one of your functions is of the form t^n where n is an integer, and you choose that as the function to differentiate then the IBP process may need numerous repetitions as you reduce the power n to get something that you can integrate directly
(iii) sometimes when you do an IBP for the first time you end up with something like I = f(x) + J where I is the original integral and J is another integral that you can't work out directly. By applying IBP to J, you may be able to express J in terms of I, which gives you an equation that you can rearrange to solve for I in terms of f(x).

Reply 2

Hasufel

Original post by mikeabc

i have done a few intergration by parts questions and I am struggling to find why some equations need to have the process done twice

any explanation would be appreciated.

as an aside note, there are things which can help in IBP, for example, reduction formulas, or, like standard integrals , knowing particular formulas for certain ones helps.

As an example, if you have

\int x^{n}exp(x)dx

the integral of this is just

e^x

multiplying an alternating series of the successive derivatives of

x^n

thusly:

\int x^{4}e^{x}dx = e^{x}(x^{4}-4x^{3}+12x^{2}-24x+24)+c

Reply 3

Hunarench95

Integration by parts formula: int(u*dv/dx) =uv-int(v*du/dx)

It's when the integral of v*du/dx is to difficult to integrate by inspection - ie it's (x^2)*(sinx) or something - so you have to use a second integration by parts to put back into the original one :smile:

Reply 4

Hasufel

that negging wasn`t me, by the way!

Reply 5

davros

Original post by Hasufel

that negging wasn`t me, by the way!

Wow - 3 negs attracted for a comprehensive answer to the OP's question. Someone must be on a downer tonight :smile:

Reply 6

Indeterminate

Original post by davros

Wow - 3 negs attracted for a comprehensive answer to the OP's question. Someone must be on a downer tonight :smile:

My reaction was something like :eek:

A very precise answer by the way :smile:

Reply 7

TenOfThem

Original post by davros

Wow - 3 negs attracted for a comprehensive answer to the OP's question. Someone must be on a downer tonight :smile:

I plussed you and, hopefully the neggers are not "worth as much as me"

xx

Reply 8

davros

Original post by Indeterminate

My reaction was something like :eek:

A very precise answer by the way :smile:

I'm up to 4 negs now!

Maybe revealing that integration by parts isn't a silver bullet that will solve all integration problems is like telling small children that the Easter Bunny doesn't exist :biggrin:

Reply 9

TenOfThem

Original post by davros

telling small children that the Easter Bunny doesn't exist

WHAT?

Well where do the hidden eggs come from then?

Reply 10

davros

Original post by TenOfThem

I plussed you and, hopefully the neggers are not "worth as much as me"

xx

much appreciated :smile:

Good job I'm not a depressive personality - I never realized an explanation of integration techniques could be so contentious!

Reply 11

mikeabc

Original post by Hunarench95

so if the integral of v*du/dx is hard to intergrate, you repeat the same process where then it will be easier to intergrate

Reply 12

davros

Original post by mikeabc

so if the integral of v*du/dx is hard to intergrate, you repeat the same process where then it will be easier to intergrate

well if your u function is x^2 for example, one application of the integration by parts formula will reduce it to a multiple of x, so you may still have a product that you need to integrate, which means re-applying the IBP formula.

If you had x^8 as a multiplier, then you might have to apply IBP 7 or 8 times to get something directly integrable - it depends on the "other" function in the original integral!