Integration of cotx (1 Viewer)

[lyounamu]

How do you integrate cot x?

 

[Aerath]

It's quite simple (I think, unless I've totally fucked it). It's like how you integrate tanx.

 

[lyounamu]

It's quite simple (I think, unless I've totally fucked it). It's like how you integrate tanx.

Hahahaha...I forgot to do that. Ironically, in the question part i), I solved how to integrate tanx. How stupid.

 

[Forbidden.]

It's quite simple (I think, unless I've totally fucked it). It's like how you integrate tanx.

oh you could do that by inspection knowing that d/dx [ln f(x)] equals f'(x)/f(x) ...

 

[lyounamu]

oh you could do that by inspection knowing that d/dx [ln f(x)] equals f'(x)/f(x) ...

Meh, he probably got it from the start anyway by looking at his working out.

 

[dood09]

oh you could do that by inspection knowing that d/dx [ln f(x)] equals f'(x)/f(x) ...

agreed.

inspection is pro

 

[vds700]

oh you could do that by inspection knowing that d/dx [ln f(x)] equals f'(x)/f(x) ...

yeah much simpler than making a substitution

 

[lolokay]

wouldn't doing this by inspection still use the same method that Aerath used, just in your head?
"cotx = cosx/sinx = u'/u. where u = sinx, so u' = cosx, so the answer must be log(sinx) + C"

 

[midifile]

wouldn't doing this by inspection still use the same method that Aerath used, just in your head?
"cotx = cosx/sinx = u'/u. where u = sinx, so u' = cosx, so the answer must be log(sinx) + C"

Yep, but then you dont have to write down all the extra lines of working

 

[lolokay]

Yep, but then you dont have to write down all the extra lines of working

well it makes it easier to post your solution on here if you do write those lines out

also, in an exam, would you just write
Int cotx.dx = log(sinx) + C [by inspection] ? or would you have to show the steps like Aerath did? (can a teacher mark you down for not including working in this question)

 

[tommykins]

just write
= int f'[x]/f[x]
= ln f[x]

 

[midifile]

well it makes it easier to post your solution on here if you do write those lines out

also, in an exam, would you just write
Int cotx.dx = log(sinx) + C [by inspection] ? or would you have to show the steps like Aerath did? (can a teacher mark you down for not including working in this question)

For logs you dont have to do write "by inspection" because they get the idea that you understand it. But you should write the line of working Int(cosx/sinx)dx. Dont just go straight from Int.cotx to the answer.

 

[lolokay]

ok, fair enough

 

[conics2008]

if u think integration of cot is hard.. try these

S sec(x)
S cot^2(x)
S cosec^2(x)

HINT : find d/dx cosec , d/dx sec, d/dx of cot

for S sec.. change to suit this f'(x)/f(x)

 

[lolokay]

sec[x] is the only hard one of those
my way of doing it would be: write sec[x] as cosec[pi/4 - x]
= Int 1/sin[pi/4 - x]
= 1/2 Int 1/sin[pi/8 - x/2]cos[pi/8 - x/2]
let u = [pi/8 - x/2]

1/2 Int (sin²u + cos²u)/(sinu cosu) .dx/du .du
dx/du = 1/(d(pi/8 - x/2)/dx) = 1/(-1/2) = -2

- Int sinu/cosu + cosu/sinu . du
= -(-log[cosu] + log[sinu])
= log[cot[pi/8 - x/2]]

 

[tommykins]

回复: Re: Integration of cotx

sec[x] is the only hard one of those
my way of doing it would be: write sec[x] as cosec[pi/4 - x]
= Int 1/sin[pi/4 - x]
= 1/2 Int 1/sin[pi/8 - x/2]cos[pi/8 - x/2]
let u = [pi/8 - x/2]

1/2 Int (sin²u + cos²u)/(sinu cosu) .dx/du .du
dx/du = 1/(d(pi/8 - x/2)/dx) = 1/(-1/2) = -2

- Int sinu/cosu + cosu/sinu . du
= -(-log[cosu] + log[sinu])
= log[cot[pi/8 - x/2]]

Wow mate, no need for such complexity.

int secx = int secx(secx + tanx)/(secx + tanx) = ln (sec x + tan x) + c

 

[lolokay]

Re: 回复: Re: Integration of cotx

but how would you know to put (sec x + tan x) in?

 

[tommykins]

回复: Re: 回复: Re: Integration of cotx

There was a proof/reason somewhere in my notes, but I can't be bothered finding it.

 

[lolokay]

Re: 回复: Re: 回复: Re: Integration of cotx

if you didn't know to do it though, I would imagine it would be a pretty complex method too - basically solving a differential equation (?) for f'(x) = f(x)/cosx

 

[Aerath]

Re: 回复: Re: Integration of cotx

Will you be required to integrate those sorta trig identities in 2U course?