Using 1 in Integration By Parts Watch

ViralRiver
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Report Thread starter 7 years ago
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How do you know when to use it? I know you use it when integrating lnx, but now I'm doing FP3 I have to do it for integrating arsinhx, arctanx etc. But I don't know exactly which ones to use it on... so is there a definitive list or a way of knowing?
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olipal
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For Further Maths A level, you will be expected to use 1x(whatever you are trying to integrate) for inverse trigonometric and hyperbolic functions, and also for ln. In each case, assuming you are going by \displaystyle\int u v' = u v - \displaystyle\int u' v, let v'=1. Setting v' as the thing you were initially asked to integrate would be a bit pointless; if you are able to find v from v' then there would be no need for using this method at all
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TwilightKnight
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(Original post by ViralRiver)
How do you know when to use it? I know you use it when integrating lnx, but now I'm doing FP3 I have to do it for integrating arsinhx, arctanx etc. But I don't know exactly which ones to use it on... so is there a definitive list or a way of knowing?
I just generally use LIATE to assign 'u' - Logarithms, Inverse(s), Algebraic, Trigonometic/Hyperbolic, Exponentials.

1 Counts as an Algebraic. If you've got an inverse, i.e, artanhx, then LIATE says to use the inverse (which is essentially a logarithm here) as u and 1 as dv/dx. Using LIATE works for any expression, so even if you didn't have to use '1', you could just use parts to integrate it, so if you want to be safe, do that. I.e, integrating tanx - if you use LIATE, you would set u=1 and dv/dx = tanx, which would give you, du/dx = 0 and v=ln(secx). If you then just plug that into the parts formula, you get the standard integral of ln(secx).

If you're asking how you're supposed to spot it in general, the general rule is that if you see something by itself, that you know doesn't integrate directly, you use u={function} and dv/dx = 1. Like you've said, in FP3 that's almost exclusively the inverses and natural logs.
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ViralRiver
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Report Thread starter 7 years ago
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Ahh ok, thanks a lot for the insight .
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