Maths Question

CookieMonster.x

I am doing an integration question from the 2013 sample paper. It's q8 c) i).
So I have gotten to f du/u is equalled to ln u. I had this done before but I used e-xamit.ie so I'm not totally sure about it.
I was wondering why it's ln u? I know that that's the integral but I thought tht was when it was 1/x, so I took the du out to get
1/u.du. I'm obviously wrong, can anyone tell me why the du counts?
I have a feeling it's something really silly, I've a headache and have been making loads of silly mistakes today :P I don't really like integration because I missed so much of it in school but I know it is easy to get marks in so I need to stick with it.
If anyone could help me out I'd really appreciate it

TheBody

CookieMonster.x wrote: »

I am doing an integration question from the 2013 sample paper. It's q8 c) i).
So I have gotten to f du/u is equalled to ln u. I had this done before but I used e-xamit.ie so I'm not totally sure about it.
I was wondering why it's ln u? I know that that's the integral but I thought tht was when it was 1/x, so I took the du out to get
1/u.du. I'm obviously wrong, can anyone tell me why the du counts?
I have a feeling it's something really silly, I've a headache and have been making loads of silly mistakes today :P I don't really like integration because I missed so much of it in school but I know it is easy to get marks in so I need to stick with it.
If anyone could help me out I'd really appreciate it

Hi there. To be honest, I'm having difficulty reading your post. Anyway, is it this question you are having trouble with?

[latex]I=\int^b_a\frac{\cos x}{1+\sin x}dx[/latex]

CookieMonster.x

TheBody wrote: »

Hi there. To be honest, I'm having difficulty reading your post. Anyway, is it this question you are having trouble with?

[latex]I=\int^b_a\frac{\cos x}{1+\sin x}dx[/latex]

Sorry I did realise I was probably a bit confusing! I can't see what you have written on my iPod but once I am on a computer I will come back to you, thank you!
I'll try repeat it a bit more coherently!
So 1/x = ln x (loge x)
In the question, I got du/u (substitution) and saw that the next step was to put it into the form ln u (loge u).
My question was why did you not have to change the du on top, because it is not equal to 1.
Hope that's a bit clearer. Thanks!

TheBody

CookieMonster.x wrote: »

Sorry I did realise I was probably a bit confusing! I can't see what you have written on my iPod but once I am on a computer I will come back to you, thank you!
I'll try repeat it a bit more coherently!
So 1/x = ln x (loge x)
In the question, I got du/u (substitution) and saw that the next step was to put it into the form ln u (loge u).
My question was why did you not have to change the du on top, because it is not equal to 1.
Hope that's a bit clearer. Thanks!

Ah ok. I think I understand now. The "missing 1" is sort of hidden. du/u=(1/u)du. I can do it all out for you if you wish. You will need to be at your computer to see the lines of maths though.

decisions

I'm really not understanding what your question is.

What I think you are having trouble with is changing from x's to u's and then integrating. You put u=x, differentiate it and then and cross multiply du/dx to get du= whatever dx.

From there sub u into what you are integrating, change your limits and I'm assuming you got 1/u du. Integrating 1/u du is the same as integrating 1/x dx.

So you get lnu and then you put in the limits and finish.

Or what some people do is instead of changing limits make them a and b, then change u to x, so you have lnx and you can then use the original limits.

CookieMonster.x

decisions wrote: »

I'm really not understanding what your question is.

What I think you are having trouble with is changing from x's to u's and then integrating. You put u=x, differentiate it and then and cross multiply du/dx to get du= whatever dx.

From there sub u into what you are integrating, change your limits and I'm assuming you got 1/u du. Integrating 1/u du is the same as integrating 1/x dx.

So you get lnu and then you put in the limits and finish.

Or what some people do is instead of changing limits make them a and b, then change u to x, so you have lnx and you can then use the original limits.

Ah I get you, I didn't realise that it was 1/x dx, I thought it was just 1/x.
Thanks for the help!

decisions

CookieMonster.x wrote: »

Ah I get you, I didn't realise that it was 1/x dx, I thought it was just 1/x.
Thanks for the help!

You only drop the dx when you integrate.

CookieMonster.x

decisions wrote: »

You only drop the dx when you integrate.

Thanks, haven't done integration in a while!