The Student Room Group
Reply 1
cosec x = 1/sinx = sinx/sin²x = (sin x)/(1 - cos²x). Let u = cos x and then proceed by partial fractions after the substitution :biggrin: .

Hope this helps!

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

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

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?
Reply 4
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.
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 :biggrin:
Reply 6
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.
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.
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.
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.
Reply 10
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
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.
Reply 12
For Edexcel C4 as it currently stands, you are expected to know that cscxdx=ln[A(cscx+cotx)],\int \csc x \:{\rm d}x= -\ln[A(\csc x + \cot x)], but you're not expected to be able to show it.
Reply 13
Actually I'm pretty sure it's in the formula book, so you're not expected to know it at all even :smile: .
Reply 14
Well, there you go. :smile:
I want to knw about integartion of cosec thita
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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Reply 17
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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