creativebuzz
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I can integrate this fine when rearranging my equation in terms of cosec, but my question is that can you integrate this using cot, and if not, why?

Because the integral of cotx is lnsinx so can't you just integrate (cot2x)^2
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ghostwalker
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(Original post by creativebuzz)
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I can integrate this fine when rearranging my equation in terms of cosec, but my question is that can you integrate this using cot, and if not, why?

Because the integral of cotx is lnsinx so can't you just integrate (cot2x)^2
How would you do it?
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creativebuzz
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(Original post by ghostwalker)
How would you do it?
I wasn't entirely sure but could you do what you would do if you were dealing with (sin2x)^2 or (cos2x)^2 etc etc by doing 1/3(lnsin2x)(1/2) + c which is 1/6(lnsin2x) + c
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Jimmy20002012
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Try using an appropriate trig identity.


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Roxanne18
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(Original post by creativebuzz)
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I can integrate this fine when rearranging my equation in terms of cosec, but my question is that can you integrate this using cot, and if not, why?

Because the integral of cotx is lnsinx so can't you just integrate (cot2x)^2
Which exam board is this from ?
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creativebuzz
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(Original post by Roxanne18)
Which exam board is this from ?
It's just an edexcel soloman paper
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creativebuzz
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(Original post by Jimmy20002012)
Try using an appropriate trig identity.


Posted from TSR Mobile
You might want to read the initial post again :P
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Roxanne18
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(Original post by creativebuzz)
It's just an edexcel soloman paper
Ah okay, so what's a soloman paper ?
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ghostwalker
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(Original post by creativebuzz)
I wasn't entirely sure but could you do what you would do if you were dealing with (sin2x)^2 or (cos2x)^2 etc etc by doing 1/3(lnsin2x)(1/2) + c which is 1/6(lnsin2x) + c
If I was dealing the (sin2x)^2 or (cos2x)^2, I'd use the double angle formula with cos4x.

Really not clear what you're proposing "1/3(lnsin2x)(1/2) + c" is the integral of.
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creativebuzz
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(Original post by ghostwalker)
If I was dealing the (sin2x)^2 or (cos2x)^2, I'd use the double angle formula with cos4x.

Really not clear what you're proposing "1/3(lnsin2x)(1/2) + c" is the integral of.

Yeah I realised that afterwards :P I think what I was mixing it up is when you can integrate (sinx)^2 by working backwards (differentiating)!

But if we can integrate sin2x and we can integrate (sinx)^2, why can't we integrate (sin2x)^2 without re-writing it?
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Jai Sandhu
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(Original post by creativebuzz)
Yeah I realised that afterwards :P I think what I was mixing it up is when you can integrate (sinx)^2 by working backwards (differentiating)!

But if we can integrate sin2x and we can integrate (sinx)^2, why can't we integrate (sin2x)^2 without re-writing it?
You cannot integrate:

\sin^2 \theta

without using a double angle formulae.
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urkadee
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(Original post by boromir9111)
Spoiler:
Show
\int cot^22x\ dx

=\int cosec^22x-1\ dx

=-\frac{1}{2}cot2x-x+C
But how do you jump to that last line? The integral of cosec(x) is -ln|cosec(x) + cot (x)

Or are you using the reverse of the differential formulae given in the C3 tables?
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boromir9111
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(Original post by urkadee)
But how do you jump to that last line? The integral of cosec(x) is -ln|cosec(x) + cot (x)

Or are you using the reverse of the differential formulae given in the C3 tables?
I deleted my post because OP said he/she knew how to do it in that form. But, since you have asked:

integral of (1/sin^2(x)) dx = integral (sinx/sin^3(x)) dx =
integral (sin(x)/(1-cos^2(x))^3/2 = integral (-du/(1-u^2)^(3/2)......this integral can be evaluated:

integral (-du/(1-u^2)^(3/2) = (-u/(1-u^2)^(1/2))+ C = -cos(x)/sinx + C
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the bear
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(Original post by Roxanne18)
Ah okay, so what's a soloman paper ?
Solomon Maths papers are hard practice papers, mainly for Edexcel and OCR ?

http://www.churchillmaths.co.uk/cmlw...mon-press.html
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urkadee
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(Original post by boromir9111)
I deleted my post because OP said he/she knew how to do it in that form. But, since you have asked:

integral of (1/sin^2(x)) dx = integral (sinx/sin^3(x)) dx =
integral (sin(x)/(1-cos^2(x))^3/2 = integral (-du/(1-u^2)^(3/2)......this integral can be evaluated:

integral (-du/(1-u^2)^(3/2) = (-u/(1-u^2)^(1/2))+ C = -cos(x)/sinx + C
Can you explain the second last line? Im getting confused by the denominator.

why does it go from (1-u^2)^3/2 to ((1-u^2)^1/2)?
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creativebuzz
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(Original post by Jai Sandhu)
You cannot integrate:

\sin^2 \theta

without using a double angle formulae.
Yeah I already mentioned that I knew that! But I was wondering why that's the case considering you can integrate sin2x and (sinx)^2, I just wanted to know the exact reasoning behind it really..
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urkadee
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Correct me if I'm wrong, but if you were to differentiate cot(2x), applying the first line in the C3 tables tan(kx) = ksec^2(kx)

This would mean that if you were to differentiate cot(2x), you would get -2cosec^2(2x), not cosec^2(2x).

So integrating -2cosec^2(2x) would get you cot2x.
You would need the 2 in "-2cosec" to get the cot2x.
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ghostwalker
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(Original post by creativebuzz)
Yeah I realised that afterwards :P I think what I was mixing it up is when you can integrate (sinx)^2 by working backwards (differentiating)!

But if we can integrate sin2x and we can integrate (sinx)^2, why can't we integrate (sin2x)^2 without re-writing it?
When you integrate (sinx)^2 you rewrite it, in a similiar fashion to (sin2x)^2
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Jai Sandhu
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(Original post by creativebuzz)
Yeah I already mentioned that I knew that! But I was wondering why that's the case considering you can integrate sin2x and (sinx)^2, I just wanted to know the exact reasoning behind it really..
 \sin \2theta is in the same general form as  \sin \theta , i.e. anything of the form  \sin {(k\theta+c)} can be integrated straight up. However, I cannot explain the exact reasoning behind why it does not work for something squared, however, by trail and error, integrate  \sin^2 \theta you way you would do  \sin \2theta and then differentiate it. You will find that when you differentiate it you do not get the same thing you started with.
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AmarPatel98
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If you were to integrate (cotx)^2, as you asked, you encounter a problem when correcting the coefficient.

By this I mean, if you were to integrate, say, cos 2x ======> 1/2 sin 2x

You write a '1/2' in front to correct the coefficient (because if you were to differentiate sin 2x you get 2 cos 2x; so multiplying this by 1/2 removes the 2 effectively)

If you tried the same thing with (cot x)^2, writing this as (ln| sin x |)^2 would not be appropriate, because if you were to differentiate this, you get 2cos x / sin x.

You have an unwanted cos x, for which you cannot 'correct' - you can only correct for integers.

Hope this helps!
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