integration of cosec(x)

speedfreak2007

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.

Reply 1

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:

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Hope this helps!

~~Simba

Reply 2

GoodbyeTSR

the answer is

-ln(cosecx+cotx)

try working backwards.

Reply 3

speedfreak2007

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

Simba

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.

Reply 5

speedfreak2007

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

Operator

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.

Reply 7

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.

Reply 8

Totally Tom

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.

Reply 9

silent ninja

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

SimonM

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

Reply 11

silent ninja

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.