A level maths integration help!!!!!!

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PeaceMinusOne
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#1
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#1
I got -1/2 instead of -1/4, due to there being an extra x2
The link to the answer is here:https://imgur.com/gallery/UJIcLIW
Can someone explain why there is an extra x2???
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flumefan1
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#2
(Original post by PeaceMinusOne)
I got -1/2 instead of -1/4, due to there being an extra x2
The link to the answer is here:https://imgur.com/gallery/UJIcLIW
Can someone explain why there is an extra x2???
Can I see your working, and the original question please?
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zetamcfc
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(Original post by PeaceMinusOne)
I got -1/2 instead of -1/4, due to there being an extra x2
The link to the answer is here:https://imgur.com/gallery/UJIcLIW
Can someone explain why there is an extra x2???
Because they have applied the chain rule twice and I suspect you have not.
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PeaceMinusOne
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(Original post by zetamcfc)
Because they have applied the chain rule twice and I suspect you have not.
I did apply the chain rule, as far as I know.
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zetamcfc
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(Original post by PeaceMinusOne)
I did apply the chain rule, as far as I know.
For the derivative of cot(2x)?
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flumefan1
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(Original post by PeaceMinusOne)
I did apply the chain rule, as far as I know.
You will need to apply it twice, not once.
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PeaceMinusOne
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(Original post by zetamcfc)
For the derivative of cot(2x)?
I did it for (cot(2x))^2

and I got, -2cosec^2(2x)cot(2x)
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PeaceMinusOne
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(Original post by flumefan1)
You will need to apply it twice, not once.
Do you know why it needs to be applied twice?
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PeaceMinusOne
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#9
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I think it has something to do with the 2x, maybe the double angle formula is involved some way?
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flumefan1
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(Original post by PeaceMinusOne)
Do you know why it needs to be applied twice?
I would be able to explain better if I knew the original question, but to find the derivative of some functions you'll need to do it twice.
Did you get my PM?
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zetamcfc
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(Original post by PeaceMinusOne)
Do you know why it needs to be applied twice?
By definition? You need to revisit the chain rule.
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PeaceMinusOne
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#12
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(Original post by flumefan1)
I would be able to explain better if I knew the original question, but to find the derivative of some functions you'll need to do it twice.
The question is 2b and I have also attached my working out.
Attached files
Last edited by PeaceMinusOne; 1 month ago
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mqb2766
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What would you get for differentiating
cot^2(u)
Last edited by mqb2766; 1 month ago
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PeaceMinusOne
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(Original post by mqb2766)
What would you get for differentiating
cot^2(x)
the answer is -2cosec^2(x)cot(x) I believe..
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mqb2766
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#15
(Original post by PeaceMinusOne)
the answer is -2cosec^2(x)cot(x) I believe..
Yes. Now let
x = 2z
What's the derivative with respect to z?
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flumefan1
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(Original post by PeaceMinusOne)
The question is 2b and I have also attached my working out.
Use substitution, where u = cot(2x), so u(1)(x) (first derivative) = -2cosec2(2x), and dx = -1/2*sin2(2x).
Substituting this in gives the integral of -0.5u, which is -u2/4.
Undoing the substitution gives -1/4* cot2(2x) + C as required.
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PeaceMinusOne
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#17
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(Original post by mqb2766)
Yes. Now let
x = 2z
What's the derivative with respect to z?
the derivative of x=2z with respect to z is just 2?
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zetamcfc
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#18
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(Original post by flumefan1)
Use substitution, where u = cot(2x), so u(1)(x) (first derivative) = -2cosec2(2x), and dx = -1/2*sin2(2x).
Substituting this in gives the integral of -0.5u, which is -u2/4.
Undoing the substitution gives -1/4* cot2(2x) + C as required.
A point on that, why does wolfram give a different answer? https://www.wolframalpha.com/input/?...E2+cot%282x%29
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Sinnoh
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#19
(Original post by PeaceMinusOne)
the derivative of x=2z with respect to z is just 2?
So you bring that derivative out to the front, where there's already a factor of 2...
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flumefan1
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#20
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(Original post by zetamcfc)
A point on that, why does wolfram give a different answer? https://www.wolframalpha.com/input/?...E2+cot%282x%29
It doesn't?
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