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speedfreak2007
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#1
Report Thread starter 12 years ago
#1
This is what I have so far, but I am aware it is wrong. Could someone tell me where it goes wrong please:

int(1/sin(x)dx
u = sin(x)
du/dx = cos(x)
dx/du = 1/cos(x)

int(1/sin(x) dx = int((1/u)du.(dx/du)) = ln(u).(1/cos(x)) = (ln(sin(x)))/cos(x)

Any help would be appreciated.
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Simba
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#2
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cosec x = 1/sinx = sinx/sin²x = (sin x)/(1 - cos²x). Let u = cos x and then proceed by partial fractions after the substitution .

Hope this helps!

~~Simba
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thomaskurian89
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#3
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the answer is -ln(cosecx+cotx)

try working backwards.
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speedfreak2007
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#4
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(Original post by Simba)
cosec x = 1/sinx = sinx/sin²x = (sin x)/(1 - cos²x). Let u = cos x and then proceed by partial fractions after the substitution .

Hope this helps!

~~Simba
Ok I think I can do it that way but I still want to know what is wrong with my method and where I went wrong? Or is there no way of telling?
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Simba
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#5
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(Original post by speedfreak2007)
Ok I think I can do it that way but I still want to know what is wrong with my method and where I went wrong? Or is there no way of telling?
The problem lies here:

int(1/sin(x) dx = int((1/u)du.(dx/du)) = ln(u).(1/cos(x))

You can't 'individually' integrate so to speak the 1/u. You need to integrate (dx/du)(1/u) with respect to u.
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speedfreak2007
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#6
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Ok thank you. In theory then, my way would never work because i can't integrate a function with both u and cos(x). You'd think i'd know this doing further maths and all...
Oh well. Thanks
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Operator
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#7
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You can also do it by substituting t = tan(x/2). Admittedly, that's not the quickest way to do it, but it's alright.
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silent ninja
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#8
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Another method is to multiply by (cosecx+cotx)/(cosecx+cotx) and go from there (it becomes f'(x)/f(x) form with adjustment). It's a shorter method.
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Totally Tom
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#9
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(Original post by silent ninja)
Another method is to multiply by (cosecx+cotx)/(cosecx+cotx) and go from there (it becomes f'(x)/f(x) form with adjustment). It's a shorter method.
but thinking to multiply by that in the first place isn't quite so obvious.
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silent ninja
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#10
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(Original post by Totally Tom)
but thinking to multiply by that in the first place isn't quite so obvious.
Nah it's not. There are a few different methods, including the t substitution mentioned above. None of them are entirely obvious I dont think.
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SimonM
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#11
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The t solution is well known though, due to its use in solving other problems so its not unreasonable to expect someone to use it
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silent ninja
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#12
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Except (for Edexcel at least) you dont meet it until FP2 and the question above could well be a C4 question? Well I think it's C4, I dont know the modules well enough anymore. T substitution was never explicitly mentioned in any of my books anyway.
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Operator
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#13
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For Edexcel C4 as it currently stands, you are expected to know that \int \csc x \:{\rm d}x= -\ln[A(\csc x + \cot x)], but you're not expected to be able to show it.
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Simba
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#14
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Actually I'm pretty sure it's in the formula book, so you're not expected to know it at all even .
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Operator
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#15
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Well, there you go.
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ahmar shah
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#16
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I want to knw about integartion of cosec thita
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_gcx
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#17
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#17
(Original post by ahmar shah)
I want to knw about integartion of cosec thita
Please create another thread for this, this one is over 9 years old!

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lock pls Sonechka Notnek
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Sir Cumference
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#18
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#18
(Original post by ahmar shah)
I want to knw about integartion of cosec thita
This is a very old thread. Please start a new one if you want to ask a question.

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