# C4 Integration Question Watch

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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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#2

(Original post by

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

**creativebuzz**)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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(Original post by

How would you do it?

**ghostwalker**)How would you do it?

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#5

**creativebuzz**)

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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(Original post by

Which exam board is this from ?

**Roxanne18**)Which exam board is this from ?

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#8

(Original post by

It's just an edexcel soloman paper

**creativebuzz**)It's just an edexcel soloman paper

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#9

(Original post by

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

**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

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

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(Original post by

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.

**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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#11

(Original post by

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)!

**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?**without using a double angle formulae.

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#12

Or are you using the reverse of the differential formulae given in the C3 tables?

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#13

(Original post by

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?

**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?

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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#14

(Original post by

Ah okay, so what's a soloman paper ?

**Roxanne18**)Ah okay, so what's a soloman paper ?

http://www.churchillmaths.co.uk/cmlw...mon-press.html

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#15

(Original post by

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

**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

why does it go from (1-u^2)^3/2 to ((1-u^2)^1/2)?

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#17

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.

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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#18

**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?**

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#19

(Original post by

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..

**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..

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#20

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!

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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