Honours Maths Question

cookiemonst3r

1996 paper 1 integration q:

integrate e^x/ 1 + e^2x
definate integrals from ln(root3) to 0

Help!!

Irish Wolf

let u = e^x

get the natural log of both sides => ln(u) = x

differentiate w.r.t du => du/u = dx

and substitute back into original integral you get integral of (u/(1+u^2))(du/u)

cancel the u above and below the line and it reduces to integral of du/(1+u^2)

integrate and substitute back for u...

it's been a long time since I did anything like this - so it could be a little fuzzy...

Nehpets

Yep, that's right. Thanks Irish Wolf. That's a bitch of a question

Irish Wolf

From what I remember its a matter of selecting the correct term to substitute so you can isolate the x term and are just left with some form of a u.du integral...

cookiemonst3r

thanks so much!! do u know the answer to it though?? i think the one in the back of the book is wrong. In the q it says to give it correct to 3 decimal places but the answer in the back of the book is pi/3.

Nehpets

Yeah, what I was trying, was to substitute the bottom..

so

u = 1 + e^2x
du = 2e^2x dx

1/2e^x du = e^x dx

doesn't work..

Irish Wolf

I haven't worked through the final integral - try it out yourself ...

BTW pi/3 can be expressed to 3 decimal places - 1.047.

cookiemonst3r

i got ln2 as my answer

MathsManiac

Above technique is correct, but you don't need to take the logs, by the way.

Once you let u=e^x, you have du=e^xdx straight away, giving the integral of du over 1+u^2.

This gives arctan(u), as I'm sure you got, yielding arctan(e^x) to be evaluated from x=0 to x=ln(sqrt(3)). I make that Pi/12, or 0.262.

Does this correspond with the given answer?

analyse this

Oh god i have spent the last half hour at it!!...and I seem to have made no progress!:( Can someone show me where I went wrong?

Integration of= Int.

Int. e^x/1+e^2x dx with limits ln√¯¯3 and 0

let u= i+e^2x
e^2x=u-1
(e^x)^2= u-1
e^x= √¯¯u-1
du/dx=2e^2x
du/dx=2(e^x)^2
du/2e^x=e^x dx
du/2√¯¯u-1= e^x dx

Int. e^x/1+e^2x= Int. 1/u X du/2√¯¯u-1 (with limits ln√¯¯3 and 0)

= 1/2 Int. du/u√¯¯u-1 (with above limits)
= 1/2 Int. (u-1)^-1/2 X 1/u du (with limits)
=1/2 Int. (u-1)^-1/2 du X Int. 1/u du (with limits)
=1/2 [2√¯¯u-1](with limits) X [log u] (with limits)
=[√¯¯u-1] (with limits) X [log u] (with limits)

This is where I get completely lost!! Please help! I hope we dont get anything as hard as this in the exam....or I am completely screwed.

cookiemonst3r

MathsManiac wrote:

Above technique is correct, but you don't need to take the logs, by the way.

Once you let u=e^x, you have du=e^xdx straight away, giving the integral of du over 1+u^2.

This gives arctan(u), as I'm sure you got, yielding arctan(e^x) to be evaluated from x=0 to x=ln(sqrt(3)). I make that Pi/12, or 0.262.

Does this correspond with the given answer?

I got that answer too. The back of the book gives pi/3 but it could be wrong

Irish Wolf

MathsManiac wrote:

Above technique is correct, but you don't need to take the logs, by the way.

Once you let u=e^x, you have du=e^xdx straight away, giving the integral of du over 1+u^2.

This gives arctan(u), as I'm sure you got, yielding arctan(e^x) to be evaluated from x=0 to x=ln(sqrt(3)). I make that Pi/12, or 0.262.

Does this correspond with the given answer?

Of course... but as I said it would probably be a little fuzzy...

So when you apply the limits you get

*edit*

arctan(root3) - arctan(1) = pi/12

ah well - the method was correct so that's most of the marks - right?

ZorbaTehZ

MathsManiac wrote:

Above technique is correct, but you don't need to take the logs, by the way.

Once you let u=e^x, you have du=e^xdx straight away, giving the integral of du over 1+u^2.

This gives arctan(u), as I'm sure you got, yielding arctan(e^x) to be evaluated from x=0 to x=ln(sqrt(3)). I make that Pi/12, or 0.262.

Does this correspond with the given answer?

Thats the way I did it too.

However the way IrishWolf did it is ingenious imo, I'd never have thought of that way tbh.

EDIT: Disregarding the discrepancies in answer ofc.

Irish Wolf

ZorbaTehZ wrote:

Thats the way I did it too.

However the way IrishWolf is ingenious imo, I'd never have thought of that way tbh.

Haha! Flattery! But its the same method and the Maniac's was much more succinct...

Nehpets

What is arctan? ...

ZorbaTehZ

Inverse Tangent (Tan^-1)

Nehpets

Oh, never knew it was called that

is the integral not:

(1/u)arctan(1/u)

ZorbaTehZ

Irish Wolf wrote:

Haha! Flattery! But its the same method and the Maniac's was much more succinct...

lal, aye, I realise now that they're almost exactly the same.:)

analyse this

Nehpets wrote:

What is arctan? ...

Pffffff...amateur!:cool:

Well I'm not gonna tell him...someone else can:rolleyes:

Only messing I have never come across the word in my life:o

MathsManiac

btw, I only ever say arctan when typing, cos it's less hassle than tan^-1

Sorry, should have explained it.:o

Nehpets

haha, I'm not getting that answer

EDIT:

Int. 1 / (1+u^2)

= (1/u)arctan(1/u)

No?

MathsManiac

analyse this wrote:

Oh god i have spent the last half hour at it!!...and I seem to have made no progress!:( Can someone show me where I went wrong?

Integration of= Int.

Int. e^x/1+e^2x dx with limits ln√¯¯3 and 0

let u= i+e^2x
e^2x=u-1
(e^x)^2= u-1
e^x= √¯¯u-1
du/dx=2e^2x
du/dx=2(e^x)^2
du/2e^x=e^x dx
du/2√¯¯u-1= e^x dx

Int. e^x/1+e^2x= Int. 1/u X du/2√¯¯u-1 (with limits ln√¯¯3 and 0)

= 1/2 Int. du/u√¯¯u-1 (with above limits)
= 1/2 Int. (u-1)^-1/2 X 1/u du (with limits)
=1/2 Int. (u-1)^-1/2 du X Int. 1/u du (with limits)
=1/2 [2√¯¯u-1](with limits) X [log u] (with limits)
=[√¯¯u-1] (with limits) X [log u] (with limits)

This is where I get completely lost!! Please help! I hope we dont get anything as hard as this in the exam....or I am completely screwed.

I hope at this stage you've spotted that where you went wrong was with your substitution in the first place.

When trying to decide what to substitute, it's always best to look at the whole expression, and watch to see whether the derivative of what you're planning on substituting is hangin' around just waitin' to get stuck to a dx!

MathsManiac

Nehpets wrote:

haha, I'm not getting that answer

EDIT:

Int. 1 / (1+u^2)

= (1/u)arctan(1/u)

No?

can't see where you're getting that from.

(See tables p.41, with a=1 and u=x)

JC 2K3

Might as well do it for the practice.....

(e^x/ 1 + e^2x).dx
u = e^x
du = e^x.dx

int((1/1 + u^2).du)
tan^-1(u)
tan^-1(e^x)

tan^-1(1) - tan^-1(root3)
pi/4 - pi/6
(6pi - 4pi)/24
=pi/12

w00t!

Nehpets, arctan = tan^-1

Nehpets

MathsManiac wrote:

can't see where you're getting that from.

(See tables p.41, with a=1 and u=x)

I was reading them wrong.. woops

mp3guy

does the fact it's:

integrate e^x/ 1 + e^2x

not mean anything?

I don't get what you've done here;

(e^x/ 1 + e^2x).dx
u = e^x
du = e^x.dx

int((1/1 + u^2).du)

ZorbaTehZ

e^2x = (e^x)^2

u = e^x
Therefore:

(e^x)^2 = u^2
Is that what you're asking?
(The u that's above the line, cancels with the du/u thats is re-entered)

mp3guy

ZorbaTehZ wrote:

e^2x = (e^x)^2

u = e^x
Therefore:

(e^x)^2 = u^2
Is that what you're asking?
(The u that's above the line, cancels with the du/u thats is re-entered)

Ah yes, thanks, I mixed up adding and multiplying indices.