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Maths, Addmaths and Statistics: Post your doubts here!

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Trigonometric Substitution is hard. I've only done I guess a couple of questions but they won't be a part of the exams surely!

i don't want to take a risk. i will do it by khan academy and then search for some practice questions with solutions over internet. after watching videos from khan academy i will do much practice so i don't regret on the day of exams. i advice u to do the same.
 
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We're kinda doing the same things now! :p
I was on khan academy right now!
All in all, I believe that these won't be part of the exams (I am going to do the trig. substitution since I know how to do the others), however these are good methods of getting out of unwanted situations during the paper.
 
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We're kinda doing the same things now! :p
I was on khan academy right now!
All in all, I believe that these won't be part of the exams (I am going to do the trig. substitution since I know how to do the others), however these are good methods of getting out of unwanted situations during the paper.

i covered almost all my syllabus on khan academy. i had no proper teacher. i only asked him whenever i m stuck as he is actually my physics teacher at school. he encouraged me to do add maths and promised to help. and alhamdulillah i have covered syllabus by myself and even started to do past papers.
 
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i always thought that integration by parts, u-substitution and trig substitution are not in syllabus as they are not in book but i recently saw a question which has to be done by u-substitution. so does it mean that others are also included(by parts and trig substitution)
The syllabus explicitly states that integration by parts is excluded.
So which question are you talking about? Can you post it?
 
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The syllabus explicitly states that integration by parts is excluded.
So which question are you talking about? Can you post it?
Well, the problem is that CIE does not always follow the syllabus, I've seen one or two instances where questions have come related to the topics but out of the syllabus criteria.
 
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The syllabus explicitly states that integration by parts is excluded.
So which question are you talking about? Can you post it?

syllabus no where says that integration by parts is excluded:
parts.png
here is question
substitution.png
it has to be done by substitution(examiner report)
 
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syllabus no where says that integration by parts is excluded:
View attachment 23510
here is question
View attachment 23511
it has to be done by substitution(examiner report)
My dear friend, there's no such thing as integration by parts or u-substitution involved in this question's solution. Actually you read the Examiner's report of part (a), which instructs us to derive the value of a. In that part, you just have to substitute the value of the gradient and the coordinates given. You actually misunderstood the word 'substitution'. The solution goes like this:Add. math.png
 
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syllabus no where says that integration by parts is excluded:
I meant to say the examiner report, not the syllabus.It's a mistake on my behalf. But read the following carefully.
It's an excerpt from the June 2009 examiner report:
Of the candidates who got part (i) correct a high proportion went on and were able to gain full marks for part (ii) by making use of their answer to part (i). There were still candidates who failed to realise the implication of the word ‘Hence’ used on the question paper and tried various incorrect attempts at integration. Those candidates that used integration by parts and obtained a correct solution were given full credit even though this method is not a syllabus requirement.
 

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Here's an excerpt from the June 2009 examiner report:
Of the candidates who got part (i) correct a high proportion went on and were able to gain full marks for part (ii) by making use of their answer to part (i). There were still candidates who failed to realise the implication of the word ‘Hence’ used on the question paper and tried various incorrect attempts at integration. Those candidates that used integration by parts and obtained a correct solution were given full credit even though this method is not a syllabus requirement.
Well, obviously there is no harm in practicing integration by parts; as I said to Suchal before that these questions will most probably NOT BE part of the paper but these methods will be useful in peculiar situations! Although I do agree with the fact that questions solely based on these methods will not be given.
 
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My dear friend, there's no such thing as integration by parts or u-substitution involved in this question's solution. Actually you read the Examiner's report of part (a), which instructs us to derive the value of a. In that part, you just have to substitute the value of the gradient and the coordinates given. You actually misunderstood the word 'substitution'. The solution goes like this:View attachment 23514

the rule which allows you to take a function as a single variable is derived from u-substitution. and by substitution it is simpler and can be done in mind. examiner report states that "Those candidates who realised the need for substitution found this an easy mark."
 
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I meant to say the examiner report, not the syllabus.It's a mistake on my behalf. But read the following carefully.
It's an excerpt from the June 2009 examiner report:
Of the candidates who got part (i) correct a high proportion went on and were able to gain full marks for part (ii) by making use of their answer to part (i). There were still candidates who failed to realise the implication of the word ‘Hence’ used on the question paper and tried various incorrect attempts at integration. Those candidates that used integration by parts and obtained a correct solution were given full credit even though this method is not a syllabus requirement.

obviously i have done these kinds of them many time when the last part has a derivative for a function and we have to add some constant to make it like that derivative so it's integral is that function. i won't do integration by parts as it is trickier.
 
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obviously i have done these kinds of them many time when the last part has a derivative for a function and we have to add some constant to make it like that derivative so it's integral is that function. i won't do integration by parts as it is trickier.
My friend, this is not Integration by Parts! I know it is easier with substitution but this can also be done directly using the Integration method easily.
 
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