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Kelumptus
I am a little bit confused about how the 'integration by parts' formula actually works. My problem is as follows: The integration by parts rule is device from the product rule -> (fg)' = f'g + fg' Then by integrating throughout you get this (using S as integral symbol): fg = S f'g dx + S fg' dx My confusion relates to the fact that this is not the integration by parts formula and it appears logical to me that it should be. The integration by parts formula goes one step further: S fg' dx = fg - S f'g dx This does not make sense to me since it is the integral of (fg)' that you want to s
I'm not sure I understand your question\[d(uv)=udv+vdu\]integrating both sides\[\int d(uv)=\int u dv+\int v du\]note that\[\int d(uv)=uv\]solve for \(\int udv\):\[\int udv=uv-\int vdu\]now where exactly are you confused?
What confuses me is that the formula seems to be used solve a problem such as find the integral of ∫x cos x, but wouldn't it be logical that the formula to do this is ∫d(uv)=∫udv+∫vdu (where u = x and v = cos x) ?
If you were trying to find the antiderivative of ∫uv' then this would make sense to me
i mean if you were trying to find ∫udv.
you are, but you should have v=x and u=cosx
sorry, x=u is right, but we call dv=cosx if you just left it as ∫d(uv)=∫udv+∫vdu that is cyclical and would not get you anywhere you need to get rid of the x which is done by calling u=x, which means du=1 dv=cosx->v=sinx then ∫udv=uv-∫udu gives\[\int x\cos xdx=x\sin x-\int\sin xdx\]which you can solve
it is important to define one function as u, and the other as dv the u is usually used for any extra x's out front, since du will reduce the order of x leaving something that can be integrated, as it did above
these are good notes http://tutorial.math.lamar.edu/Classes/CalcII/IntegrationByParts.aspx
Paul's online notes? Yeah i read them a lot =)
that's good, but I guess that means they won't help you answer your query.
So... is integration by parts just integrating one part? Meaning that there is a second step required?
there can be many steps, if you had\[\int x^5\sin xdx\]you would have to do it 5 times, which is where the tabular method is useful the idea is to use the formula to make the power of x smaller by calling it u, then when it appears as du in the next integral it is lower order, until it goes away entirely and you only have to integrate the dv term
check out example 9 in the link I gave you to see what I mean by the tabular method
I have to eat dinner now, hope I helped a little
Ok thanks for your help =)
welcome, feel free to bump this question so others can continue to help you :)
To clarify, substituting ∫uv' = uv - ∫ vd' for ∫xsin x makes no sense to me because ∫ x sin x is not equal to ∫u cos x (i.e. ∫uv')
" ∫ x sin x is not equal to ∫uv' " it is though, if u =x and v = -cosx
Ahh... it makes sense now. I think i was having a bit of brain fart. It seems quite obvious and logical now that you have clarified this for me. Thanks a lot. This has been bothering me all afternoon =)