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Integration by parts

Hi!
I’m not sure how you would go about this question and the mark scheme isn’t any help. 90C634C3-F613-43C5-9054-6D1283F8CDB8.jpg.jpeg
If you use LATE you should take sin2x as ‘u’ and then integrate e^cosx but apparently that’s not possible to integrate? The final answer is 2.

Thanks :smile:

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Original post by Piza
Hi!
I’m not sure how you would go about this question and the mark scheme isn’t any help. 90C634C3-F613-43C5-9054-6D1283F8CDB8.jpg.jpeg
If you use LATE you should take sin2x as ‘u’ and then integrate e^cosx but apparently that’s not possible to integrate? The final answer is 2.

Thanks :smile:

It’s not possible so you might try the setup the other way for u and v’


P.S. I would use double angle identity here.
(edited 3 years ago)
Reply 2
Original post by RDKGames
It’s not possible so you might try the setup the other way for u and v’


P.S. I would use double angle identity here.


ahhhh okay i’ll try that thanks!
You can also use the DI method, it is a very quick, efficient way of doing integration by parts :smile:
Original post by Joshwoods01
You can also use the DI method, it is a very quick, efficient way of doing integration by parts :smile:

elaborate?
1617842332378928183146.jpg
I'll use this as an example


choose something to integrate and differentiate and create a table like so including the plus and minus signs

keep differentiating and integrating until either:
1. the differential is zero
2.you can easily integrate the product of the horizontal
3. You arrive at an almost identical point as the question (like my example, the derivative has e^2x and the integral has sin(x)

take the product of all the diagonal and add them together

take the integral of the product of the horizontal, you will end up having a term that is equal to the original question but is scaled by some value, bring it all to one side and divide by that scalar, you have your answer, then plug in your bounds, with your question, differentiate e^cos x and integrate sin(2x)
Reply 6
Original post by Joshwoods01
1617842332378928183146.jpg
I'll use this as an example


choose something to integrate and differentiate and create a table like so including the plus and minus signs

keep differentiating and integrating until either:
1. the differential is zero
2.you can easily integrate the product of the horizontal
3. You arrive at an almost identical point as the question (like my example, the derivative has e^2x and the integral has sin(x)

take the product of all the diagonal and add them together

take the integral of the product of the horizontal, you will end up having a term that is equal to the original question but is scaled by some value, bring it all to one side and divide by that scalar, you have your answer, then plug in your bounds, with your question, differentiate e^cos x and integrate sin(2x)


wowww never seen that method before, thanks I’ll try it out!
Original post by Piza
wowww never seen that method before, thanks I’ll try it out!

It may look quite long winded but once you start to use it, it becomes so quick! Can save you a good few minutes in the exam (just mention that that is the method you're using!)
Reply 8
Original post by Joshwoods01
It may look quite long winded but once you start to use it, it becomes so quick! Can save you a good few minutes in the exam (just mention that that is the method you're using!)


ofc thank you!! :smile:
Original post by Piza
ofc thank you!! :smile:


no problem, let me know if you're able to solve your problem :smile:
Original post by Joshwoods01
You can also use the DI method, it is a very quick, efficient way of doing integration by parts :smile:

I don't see the relevance, honestly. DI is just a slightly more compact way of writing out integration by parts: if you can't work out how to do it the usual way, using DI isn't going to help.

The key observation here is:

Original post by Piza
ahhhh okay i’ll try that thanks!

Using the double angle identity is key here. You need to group something with ecosxe^{\cos x} in order to get something you can integrate by recognition (or by substitution, but if you can't recognize it, you probably won't be able to spot what to do).
Original post by DFranklin
Using the double angle identity is key here. You need to group something with ecosxe^{\cos x} in order to get something you can integrate by recognition (or by substitution, but if you can't recognize it, you probably won't be able to spot what to do).

It was just a personal preference for me, yeah it's essentially the same thing, I just liked this method better as I always got confused when I had to integrate 2 functions which have infinite integrands and differentials
Reply 12
Original post by Piza
Hi!
I’m not sure how you would go about this question and the mark scheme isn’t any help. 90C634C3-F613-43C5-9054-6D1283F8CDB8.jpg.jpeg
If you use LATE you should take sin2x as ‘u’ and then integrate e^cosx but apparently that’s not possible to integrate? The final answer is 2.

Thanks :smile:

Is this a real exam question or somebody's made-up question? Any "normal" person would probably replace sin2x with 2sinxcosx and then sub u = cos x, leaving a straightforward application of IBP to the 'u'-integral, rather than trying to do IBP on the initial integral :smile:
Reply 13
Original post by davros
Is this a real exam question or somebody's made-up question? Any "normal" person would probably replace sin2x with 2sinxcosx and then sub u = cos x, leaving a straightforward application of IBP to the 'u'-integral, rather than trying to do IBP on the initial integral :smile:

It looks like a question from Madasmaths (created by the TSR user TeeEm). Except for substitution, exam questions rarely force you to use a certain integration method.
Reply 14
sorry the reply took so long, but I ended up using both integration by parts and substitution because I kept ending up at dead ends by just doing it with parts. Here’s my working I subbed in the values and got the correct answer of two. (there should be an extra ‘u’ i just missed it out in a line.
46FDFBC4-9970-4FED-888A-343FCE8FF2E3.jpg.jpeg
Haven’t tried the DI method yet but i’m sure I will before exams haha
(edited 3 years ago)
Reply 15
Original post by Notnek
It looks like a question from Madasmaths (created by the TSR user TeeEm). Except for substitution, exam questions rarely force you to use a certain integration method.


yeah it’s from MadAsMaths, never really seen a qs like that before tbh
Original post by davros
Is this a real exam question or somebody's made-up question? Any "normal" person would probably replace sin2x with 2sinxcosx and then sub u = cos x, leaving a straightforward application of IBP to the 'u'-integral, rather than trying to do IBP on the initial integral :smile:

Obviously I'm not "normal"... :smile:

[I was looking at this as analogous to integrating x^3 exp(-x^2) by parts and would have done it directly ].
Reply 17
Original post by Piza
sorry the reply took so long, but I ended up using both integration by parts and substitution because I kept ending up at dead ends by just doing it with parts. Here’s my working I subbed in the values and got the correct answer of two. (there should be an extra ‘u’ i just missed it out in a line.
46FDFBC4-9970-4FED-888A-343FCE8FF2E3.jpg.jpeg
Haven’t tried the DI method yet but i’m sure I will before exams haha

Don't forget it's a definite integral - you can convert the x limits to u-limits so you can put these in directly when you have the u-integral evaluated and then you never need to write out the full version in terms of x at the end :smile:
Reply 18
Original post by DFranklin
Obviously I'm not "normal"... :smile:

[I was looking at this as analogous to integrating x^3 exp(-x^2) by parts and would have done it directly ].

Probably not much in it to be fair. I see too many people on TSR trying to "recognize" things that's aren't true and then making a horrible mess of their integration, so if I'm putting forward a method myself here I'd always try being more explicit about what's going on, at the risk of being slightly inefficient.

Mind you, I have no idea what "LATE" and "DI" methods are supposed to be - there seems to be a modern tendency to pile methods on top of methods for the sake of having an acronym, instead of just following the basic principle of IBP. Apart from slightly more subtle examples where you use "1" as one of the functions, there are usually only 2 choices for what to differentiate vs what to integrate, and you pretty soon find out if you've made the wrong one :biggrin:
Original post by davros
Mind you, I have no idea what "LATE" and "DI" methods are supposed to be - there seems to be a modern tendency to pile methods on top of methods for the sake of having an acronym, instead of just following the basic principle of IBP.

FWIW, I suspect LATE is "log, arccos/sin/tan, trig, exp" (in terms of things you probably want to differentiate rather than integrate when doing IBP). And the "DI method" is writing two columns for the u, v parts in IBP and then successively differentiating one column and integrating the other, then with a suitable combining of entries you can write down the result of multiple IBP steps.

Although like you I've never used them, so I may be talking nonsense! :smile:

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