First Order Differentiation Prolem Watch

studienka
Badges: 1
Rep:
?
#1
Report Thread starter 11 years ago
#1
As I don't like latex - I've put the question in a word file.

I'm a bit lost as to why I'm having a problem here - presumably I'm meant to use a double angle formula somewhere to get the x to x/2, but I can't figure out when.

Hope someone can help
Attached files
0
quote
reply
Chewwy
Badges: 15
Rep:
?
#2
Report 11 years ago
#2
differentiate?
0
quote
reply
insparato
Badges: 13
Rep:
?
#3
Report 11 years ago
#3
You need to change

 -ln|cosecx + cotx| (integral of cosecx) into something that looks like the answer they give. (If you look in the edexcel formula book, its a standard manipulation of this integral)

and then apply the conditions given.

I dont understand where you're getting tan 90 from.
0
quote
reply
studienka
Badges: 1
Rep:
?
#4
Report Thread starter 11 years ago
#4
yeah - I know that's what I need to change. I just don't know how though
0
quote
reply
insparato
Badges: 13
Rep:
?
#5
Report 11 years ago
#5
Have a go..... Post up your attempt.
0
quote
reply
studienka
Badges: 1
Rep:
?
#6
Report Thread starter 11 years ago
#6
It takes me about 10 mins to mess around with equation editor - I'm working on a mac, and everything takes 10x as long.

Will put up my workings soon
0
quote
reply
insparato
Badges: 13
Rep:
?
#7
Report 11 years ago
#7
You could try using latex.

http://thestudentroom.co.uk/showthread.php?t=403989
0
quote
reply
studienka
Badges: 1
Rep:
?
#8
Report Thread starter 11 years ago
#8
I've just come back to working on this one.

insparato - I'm exactly at the stage where I've got:
-e^-y=-ln(cosecx +cotx) + c

and I agree that I need to mess around with the ln part.

But through the notes that I've created - nothing has worked, and putting all this mess into equation editor would take too long.

If I split them apart, I'm still unsure how to get from x to x/2 as no formula I can remember appears to help me.
0
quote
reply
Dirac Delta Function
Badges: 12
Rep:
?
#9
Report 11 years ago
#9
I can see where you would be getting tan(pi/2), but it's in the form cot(pi/2), hence is well defined.

I suggest you make the effort to learn latex, it is worth it in the long run.
0
quote
reply
insparato
Badges: 13
Rep:
?
#10
Report 11 years ago
#10
-ln |cosecx + cotx|

Look at the trig functions inside the log. Try rearranging (cosecx + cotx) and dont forget theres a minus sign outside the log.

I'm not asking you put this all through your equation editor. You're stuck on a specific part, rearranging  -ln (cosecx + cotx)  to  ln (tan \frac{x}{2})

Start messing around with trig functions. sinx and cosx can be changed into half angle formula....

If you dont want to use latex, you can still adequately communciate the mathematics by using brackets and +,-,/,x,^
0
quote
reply
Dirac Delta Function
Badges: 12
Rep:
?
#11
Report 11 years ago
#11
(Original post by studienka)
I've just come back to working on this one.

insparato - I'm exactly at the stage where I've got:
-e^-y=-ln(cosecx +cotx) + c

and I agree that I need to mess around with the ln part.

But through the notes that I've created - nothing has worked, and putting all this mess into equation editor would take too long.

If I split them apart, I'm still unsure how to get from x to x/2 as no formula I can remember appears to help me.
[inspirato, sorry about mowing on your lawn, I happen to be online].

plus y=0, x = pi/2 into

e^-y=ln(cosecx +cotx) + d

you get 1 = ln(1) + d so d = 1.

hence 1-e^-y = -ln|cosecx + cotx|

what can you do with the -1 infront of the ln?
have you seen the identities:

tan x = 2t/(1-t^2)
sin x = 2t/(1+t^2)
cos x = (1-t^2)/(1+t^2)

where t = tan(x/2)
0
quote
reply
studienka
Badges: 1
Rep:
?
#12
Report Thread starter 11 years ago
#12
I've not come across those identites b4 i don't think - where are they from?

I'm happy with everything up to this point though. Are the identities derivations from C3 double angles?

I do need to learn Latex though, I didn't find the thread on it that helpful though.
0
quote
reply
insparato
Badges: 13
Rep:
?
#13
Report 11 years ago
#13
No probs Dirac Delta Function, incidentally ive just been doing some work on the dirac delta function myself .

 2sinxcosx = sin2x

What happens when you half the angle?

(1+cosx) again try looking at the cosine double angle formula specifically  cos2x = 2cos^2x - 1

I dont think he will have come across the t substitution formula Dirac Delta Function.

The threads only an introduction there is more on the wiki page which i believe is linked to the thread, trig formula can be simple put into the latex tags, like sinx ... cosx..

[latex] sinx [/latex]

You could have a bash about using the preview post button.
0
quote
reply
Dirac Delta Function
Badges: 12
Rep:
?
#14
Report 11 years ago
#14
Im not familiar with the modern A level syllibus, but they were there when I was doing A levels.

you get them from the double angle formulae:

tan(2A) = 2tan(A)/(1-tan^2(A))
sin(2A) = 2sin(A)cos(A) 

= 2tan(A)cos^2(A) 

= 2tan(A)/sec^2(A) 

= 2tan(A)/(1+tan^2(A))

and cos is just one divided by the other.
0
quote
reply
Dirac Delta Function
Badges: 12
Rep:
?
#15
Report 11 years ago
#15
(Original post by insparato)
No probs Dirac Delta Function, incidentally ive just been doing some work on the dirac delta function myself .

 2sinxcosx = sin2x

What happens when you half the angle?

(1+cosx) again try looking at the cosine double angle formula specifically  cos2x = 2cos^2x - 1

I dont think he will have come across the t substitution formula Dirac Delta Function.
Aren't you still at school?? kids these days, eh, doing advanced maths well ahead of their time, what is society coming to...
0
quote
reply
insparato
Badges: 13
Rep:
?
#16
Report 11 years ago
#16
(Original post by Dirac Delta Function)
Aren't you still at school?? kids these days, eh, doing advanced maths well ahead of their time, what is society coming to...
Ha, Well ive finished school see just waiting for them a level results, so I've been doing Laplace transforms


Back on topic,

rearrange cosecx + cotx,  2sin\frac{x}{2}cos\frac{x}{2} = sinx So what does 1+cosx equal ? using the cosine double angle formula. Infact (1+cos2x) = cos^2x is nice thing to remember for integration purposes. Dont forget about the minus sign outside the log.

If you're still not sure where its leading, ill post up what you should get so you can learn from it . Honestly its one of them things you see and remember, i remember asking the same question about a year ago on here.
0
quote
reply
studienka
Badges: 1
Rep:
?
#17
Report Thread starter 11 years ago
#17
Thnx for all the help - I've got it sorted out now.

This was a much harder question than the rest of the exercise - I don't remember those trig formula's so well.
0
quote
reply
insparato
Badges: 13
Rep:
?
#18
Report 11 years ago
#18
Okay just as long as you do understand what is going on. You need to know them trig like the back of your hand for FP1.
0
quote
reply
Mitch87
Badges: 0
Rep:
?
#19
Report 11 years ago
#19
Have you seen those identities involving 't' that Dirac Delta Function posted earlier....? If not,  \sin x can be written as  \frac{2t}{1+t^2} , where  t = \tan \frac{x}{2} and hence  cosec x = \frac{1+t^2}{2t}

If we differentiate  t = \tan\frac{x}{2} with respect to t, then we get  \frac{dt}{dt} = \frac{1}{2}sec^2\frac{x}{2} \frac{dx}{dt}

Remembering that  sec^2 x = 1 + \tan^2 x :

 1 = \frac{1}{2}sec^2\frac{x}{2} \frac{dx}{dt}  \rightarrow  dx = \frac{2}{1+t^2} dt

Thus we have,
 \int cosec x dx = \int \frac{1+t^2}{2t} . \frac{2}{1+t^2}dt = \int \frac{1}{t} dt = ln|t| + c

From our defenition of t earlier, we have  \int cosec x dx = ln|tan\frac{x}{2}| (+c) as required.
0
quote
reply
X

Reply to thread

Attached files
Write a reply...
Reply
new posts
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • University of Lincoln
    Brayford Campus Undergraduate
    Wed, 12 Dec '18
  • Bournemouth University
    Midwifery Open Day at Portsmouth Campus Undergraduate
    Wed, 12 Dec '18
  • Buckinghamshire New University
    All undergraduate Undergraduate
    Wed, 12 Dec '18

Do you like exams?

Yes (186)
18.88%
No (594)
60.3%
Not really bothered about them (205)
20.81%

Watched Threads

View All