Can't Integrate (restated)

[Roark]

Here's the problem:

 

Find the length of an arc of a the curve y = ln x between the points where x=1 and x=2. (To carry out the integration rationalise the numerator.)

 

So: y=ln x, y'=1/x

 

Length = I(sqrt(1+( 1/x^2) )dx

 

(I for integral...how do you get math symbols?)

 

"...rationalise the numerator...":

 

sqrt(1+( 1/x^2)) =

 

x^2 + 1

--------------------

x sqrt(x^2 + 1 )

 

or

 

x^2 + 1

---------------------

sqrt(x^4 + x^2 )

 

Answer:

 

sqrt(5)-sqrt(2) + ln ( ( sqrt(5) - 1 ) / 2(sqtr(2) - 1) )

 

 

How do you integrate it? Is there a standard form? From the answer it looks like they used:

 

I(sqrt(u^2 + a^2)du = 1/2 u sqrt(u^2 + a^2) + 1/2 a^2 ln (sqrt(u^2 + a^2))

 

Thanks!

[blike]

This looks like a problem for faf..

 

I'm still learning basic laws for differentiation

[Dave]

[edit: btw, i have no idea whether this is right]

 

i've had a go at it, and it seems approachable.

 

what i did was to get your integral,

 

:lint: sqrt(1+x^-2)dx

 

then apply the subsitution:

 

x = sinh(t)

=> dx = cosh(t)dt

 

and then the integral becomes

 

:lint: coth(t)cosh(t)dt

 

which can be further simplified down to

 

:lint: cosh^2(t)/sinh(t)dt

 

then by making the substitution u = cosh(t), it then becomes

 

:lint: u^2/(u^2-1)du

 

then by dividing it out you get:

 

s = u + artanh(u)

 

and if you put the limits in i *think* it should work. i could be wrong however, but it looks like you'll get the right kind of answer, since artanh(u) can be written in log form.

 

i've no idea how to do this without hyperbolic functions either

 

hope this helps.

[Dave]

i was looking at this and just saw the bit about rationalising the numerator, which was a bit silly of me, but oh well.

 

when you rationalize the numerator, you will get (as you say):

 

(x^2+1)/(x*sqrt(x^2+1))

 

but you can split this up so that it becomes:

 

:lint: (x/sqrt(x^2+1) + 1/(x*sqrt(x^2+1))dx

 

the first bit is trivial (just substitute u^2 = x^2 + 1), and then the second bit is a bit more tricky, but i suspect a trig substitution like x=tan(t) might do it.

 

well, you've got 2 separate approaches to doing it now at least

[JoeDaWolf]

i've been trying this...

 

i think in class, we would just use the calculator

 

i can't find the right parameters to use partial fractions

 

i can't use trigonmetric substituition

 

However, the fact that you have that exact answer, which checks out with the calc, means that there's probably some way to do it...hmmmmmm

 

~Wolf

[Dave]

Originally posted by JoeDaWolf

i've been trying this...

 

i think in class, we would just use the calculator

 

i can't find the right parameters to use partial fractions

 

i can't use trigonmetric substituition

 

However, the fact that you have that exact answer, which checks out with the calc, means that there's probably some way to do it...hmmmmmm

 

~Wolf

 

It should come out from the method I've stated, but I may have made a silly error somewhere. If I have time later, I'll work through it.

[zakfab]

So: y=ln x, y'=1/x

 

Length = I(sqrt(1+( 1/x^2) )dx

 

Could someone please explain how to get the formula for the length of the arc please?

[Dave]

Imagine you have some function y = f(x). Say the arc length up to a certain point on the curve is s. Then if you increment x by a small value, say delta x, then you're going to get a small increase in the arc length, delta s. Now at this point, you can form a right angled triangle, with the vertical edge being delta y, the horizonatal being delta x and the hypotenuse delta s.

 

Now by applying simple Pythagoras, (delta s)^2 = (delta x)^2 + (delta y)^2

 

Therefore by dividing through by (delta x)^2 and as delta x -> 0,

 

ds/dx = sqrt(1+(dy/dx)^2) => s = :int: sqrt(1+(dy/dx)^2) dx

 

Hope this makes sense. A diagram will clear things up if not.

 

(btw, this can be used parametrically as well: instead of dividing through by delta x, you can divide through by some other parameter delta t.)

[zakfab]

Thanks I get it now, but y = ln x is not a straight line, so you shouldn't use pythagaros. Are you just using it because it's nearly straight, or is there another way to answer the question?

[Dave]

Okay, basically, you have to imagine that you're going to take one small segment of the arc, delta s. Because it's so small, you can approximate it to be a straight line and hence you can derive it using Pythagoras.

 

A diagram would help, but unfortunately I can't post one at the moment. Hope this helps.

[zakfab]

I understand, thanks

[Dave]

I think its about time I actually posted an answer to this thread, being that it's been open for absolutely ages

 

Okay, so you have those two integrals above, namely:

 

:int: (x/sqrt(x^2+1) + 1/(x*sqrt(x^2+1))dx

 

First one, as I said, is trivial by using the substitution u^2 = x^2 + 1. That comes out as sqrt(5) - sqrt(2).

 

The other one is a little harder, but with some thought, it's not too hard. The first thing that struck me was trig substitutions. The x^2+1 sticks out like a sore thumb and it's just crying out for a trig substitution. The correct one in this case is x = tan(t).

 

Notice that (tan(t))^2 + 1 = (sec(t))^2.

 

The bounds of integration are x=1 and x=2, so therefore the new bounds with this substitution are t=arctan(2) and t=pi/4.

 

Now when you use this substitution, it will all cancel down (quite nicely) to:

 

:int: cosec(t) dt

 

Which, of course, is a standard integral. If you want to do it by hand, you need to use the subsitution u = tan(t/2). It should cancel down to the integral of cosec(t).

 

I'll post the numeric answer a bit later, but I'm a bit busy atm.