[0range]

Just tried integrating ln|cos(t)|

I did it by parts and at the end ended up with

the integral of ln|cos(t)| = the integral of ln|cos(t)|

Is this right?

[ghostwalker]

(Original post by 0range)
Just tried integrating ln|cos(t)|

I did it by parts and at the end ended up with

the integral of ln|cos(t)| = the integral of ln|cos(t)|

Is this right?

You're back where you started, so no, that's not achieved anything.

Wolfram comes up with something seriously nasty looking, which I wouldn't expect to see at A-level, so, are you sure you have the question correct. And is that the whole question?

[f1mad]

(Original post by 0range)
Just tried integrating ln|cos(t)|

I did it by parts and at the end ended up with

the integral of ln|cos(t)| = the integral of ln|cos(t)|

Is this right?

[Intriguing Alias]

(Original post by 0range)
Just tried integrating ln|cos(t)|

I did it by parts and at the end ended up with

the integral of ln|cos(t)| = the integral of ln|cos(t)|

Is this right?

Well that is right but it's not very useful. Are you sure you were meant to be integrating ln(cos(t))?



EDIT: Thought I was getting somewhere then but I hit a brick wall at integrating xtanx

[raheem94]

(Original post by 0range)
Just tried integrating ln|cos(t)|

I did it by parts and at the end ended up with

the integral of ln|cos(t)| = the integral of ln|cos(t)|

Is this right?

Just checked with wolfram alpha and it gives a very long answer involving imaginary number, click the link to see the answer, http://www.wolframalpha.com/input/?i...e+ln%28cost%29.

I am sure your question is wrong, it isn't A-Level stuff.

[0range]

(Original post by ghostwalker)
You're back where you started, so no, that's not achieved anything.

Wolfram comes up with something seriously nasty looking, which I wouldn't expect to see at A-level, so, are you sure you have the question correct. And is that the whole question?

nah it's not a question an actual question well it is kinda but I just skipped the volume of revolution part of it just trying it for fun :P

I did it like as soon as I got to the second part when I had to do parts for the third term I got an eqaution and just rearranged

I mean like
I got

Integral of ln|cos(t)| = t|lncos(t)| - t|lncos(t)| + Integral of ln|cos(t)|

So I got them equaling each other

[Intriguing Alias]

(Original post by 0range)
nah it's not a question an actual question well it is kinda but I just skipped the volume of revolution part of it just trying it for fun :P

I did it like as soon as I got to the second part when I had to do parts for the third term I got an eqaution and just rearranged

I mean like
I got

Integral of ln|cos(t)| = t|lncos(t)| - t|lncos(t)| + Integral of ln|cos(t)|

So I got them equaling each other

Obviously that's of no use, though.

But yeah it's not something you'll be successful at integrating, I'm afraid.

[Manitude]

(Original post by 0range)
nah it's not a question an actual question well it is kinda but I just skipped the volume of revolution part of it just trying it for fun :P

I did it like as soon as I got to the second part when I had to do parts for the third term I got an eqaution and just rearranged

I mean like
I got

Integral of ln|cos(t)| = t|lncos(t)| - t|lncos(t)| + Integral of ln|cos(t)|

So I got them equaling each other

Something like the above can often be useful, if you apply a process to an integral and end up with something in terms of the original integral then you can re-arrange and implicitly find the solution. It's handy for functions involving infinitely differential functions like exponential and sine or cosine.

So maybe check your working and see if you do get t|lncos(t)| - t|lncos(t)|.
Although the wolfram alpha result is disgusting so it might be an incorrectly stated question. <<<might have helped if I'd actually read all of the post I'm quoting!

[elldeegee]

(Original post by hassi94)
Well that is right but it's not very useful. Are you sure you were meant to be integrating ln(cos(t))?



EDIT: Thought I was getting somewhere then but I hit a brick wall at integrating xtanx

I'm prob being really stupid, but how is integrating ln |whatever| different to integrating ln (whatever)?

[elementosaurus]

Surely integrating ln|cos(t)| goes too...

-sin(t) / cos(t)

Which is equal too...

-tan(t)

??????????

Edit: Just noticed your question... as the other guys said, are you sure it's not meant to be integrate ln cos(t) ??? In which case I think my answer is right ^ ^

[TGH1]

(Original post by elldeegee)
I'm prob being really stupid, but how is integrating ln |whatever| different to integrating ln (whatever)?

lnx only provides real solutions for x>0, so the modulus ensures that the natural log function can give a real answer.

[0range]

Ahh right guys thanks
I'm trying something else atm but I'm kinda going very very wrong


How would you integrate e^(ln|cos(t)|)?
I mean I know you can just cancel down and then integrate sin(t) but how would you do it by the chain rule? I somehow started dealing with -cot :l

(Original post by elldeegee)
I'm prob being really stupid, but how is integrating ln |whatever| different to integrating ln (whatever)?

That's just to make sure the inside is positive (just taking the modulas) because you cant log a negative number

[SubAtomic]

(Original post by elldeegee)
I'm prob being really stupid, but how is integrating ln |whatever| different to integrating ln (whatever)?

Modulus has no negative

[TGH1]

(Original post by elementosaurus)
Surely integrating ln|cos(t)| goes too...

-sin(t) / cos(t)

Which is equal too...

-tan(t)

??????????

No, that's when you differentiate.

[Intriguing Alias]

(Original post by elldeegee)
I'm prob being really stupid, but how is integrating ln |whatever| different to integrating ln (whatever)?

Sorry I wasn't trying to pick on the brackets/modulus. I don't think it makes a difference when integrating until you give limits, and it certainly doesn't make any difference at A-level

Oh and if you just meant what do |these| mean it just means it'll take the magnitude of the domain rather than the actual value.

[porkstein]

(Original post by 0range)
How would you integrate e^(ln|cos(t)|)? I mean I know you can just cancel down and then integrate sin(t) but how would you do it by the chain rule?

Are you sure you mean chain rule? In this instance I can't see how the chain rule would be relevant. The way to do this question is to cancel down, it's as simple as that. You could try a substitution, but it's likely that throughout your calculations you'd be carrying around terms that simplify, and it makes no sense to keep them in their more complicated form. You would just have to wade through a load of working and you'd probably end up with 'integral of cos(t)' - essentially you would just take a huge detour to get to your answer. It would be akin to, whilst solving an equation, randomly inserting lines of working where you add x to both sides and then take away x from both sides - it makes no sense.

[elldeegee]

To everyone who quoted me:

haha, i know what the modulus "brackets" meant, but i thought that hassi94 was saying that there was a specific way to integrate something with those brackets as opposed to when normal brackets were used. If that had been the case i would have been surprised because i have never come across it before.

However, i now know differently
which is a relief

[TheJ0ker]

(Original post by elementosaurus)
Surely integrating ln|cos(t)| goes too...

-sin(t) / cos(t)

Which is equal too...

-tan(t)

??????????

Edit: Just noticed your question... as the other guys said, are you sure it's not meant to be integrate ln cos(t) ??? In which case I think my answer is right ^ ^

Probably should edit that again. You have differentiated the function as someone has already said.