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Quick integration question

xfootiecrazeesarax

I seem to have forgot some basic integration stuff again, hitting a blank on how to integrate this at the moment:

\int \frac{1}{(\frac{1}{r}-\frac{1}{R})^\frac{1}{2}}dr

Many thanks in advance

Reply 1

Mad Sentinel

Are you sure that's what you have to integrate? The answer is pretty messy: https://www.wolframalpha.com/input/?i=integrate+((1%2Fr)+-+(1%2FR))%5E-0.5+with+respect+to+r

Reply 2

DFranklin

Original post by Mad Sentinel

Are you sure that's what you have to integrate? The answer is pretty messy: https://www.wolframalpha.com/input/?i=integrate+((1%2Fr)+-+(1%2FR))%5E-0.5+with+respect+to+r

Wolfram is sometimes a bit wierd in the way it does integrals. If you multiply top and bottom by sqrt(r) (which obviously leaves the integral unchanged) you get quite a different answer from Wolfram.

Anyhow, I would try a substitution r = x^2 to start I think...

Reply 3

Mad Sentinel

I did as you suggested, seems to give the same answer: https://www.wolframalpha.com/input/?i=integrate+r%5E0.5%2F(r%5E0.5((1%2Fr)+-+(1%2FR))%5E0.5)+with+respect+to+r

Are you sure you didn't make a mistake when entering the formula?

Reply 4

DFranklin

Original post by Mad Sentinel

The integral is:

\dfrac{1}{\left( \frac{1}{r} - \frac{1}{A} \right)^{1/2}}

Multiply top and bottom by sqrt(r), and let's use a square root instead of ^(1/2) while we're at it.

\dfrac{\sqrt{r}}{\sqrt{r} \sqrt{\frac{1}{r} - \frac{1}{A} }}

Multiply out the denominator:

\dfrac{\sqrt{r}}{\sqrt{ 1 - \frac{r}{A} }}

We haven't changed the actual function, but plug this into Wolfram and you get an result involving arctan.

For more fun and games, let B = 1/A (it's just a constant), and rewrite as:

\dfrac{\sqrt{r}}{\sqrt{1-rB}}

Now Wolfram gives you a result involving arcsin.

Reply 5

xfootiecrazeesarax

solution:
I'm still stuck

gollygosh.png10.7KB

Reply 6

DFranklin

Original post by xfootiecrazeesarax

solution:
I'm still stuck

OK, first get rid of R from the integral by the substitution u = r / R.
Next get rid of the square roots by the substitution v = sqrt(u).
Finally, make the substitution v = sin t.

Reply 7

Mad Sentinel

Triple substitution?! It's ingenious if it works, but how do you even think to try something like that? Is it just pattern recognition or is there some sort of logical process that you follow?

Reply 8

Mad Sentinel

My maths is very rusty, but does the first substitution actually remove R from the integral? Don't we have to make it (R^-2)^-0.5 to get it inside the brackets, at which point it isn't cancelling out the other Rs?

(edited 7 years ago)

Reply 9

Zacken

FWIW, when we had to do this on our Dyn&Rel sheet 1, I used the substitution r = R sin^2 t (which is precisely what DFrank is suggesting)

Reply 10

DFranklin

Original post by Mad Sentinel

Triple substitution?! It's ingenious if it works, but how do you even think to try something like that? Is it just pattern recognition or is there some sort of logical process that you follow?

I'm kind of surprised given what I posted that you can't see it's a series of logical steps. The first substitution is essentially cosmetic, but integrating

(1/u-1)^{1/2}

just "feels" more pleasant than integrating

(1/r-1/R)^{1/2}

. It also gives us the

R^{3/2}

factor, which is reassuring.

I could have explained the 2nd sub better. I rearranged the integral to be

\dfrac{\sqrt{u}}{\sqrt{1-u}}

similarly to what I'd said in a previous post. At this point, v^2 = u gets rid of the sqrt(u) (it replaces it with something else, but it gets rid of the sqrt). The bottom becomes

\sqrt{1-v^2}

At this point you have "poly in v" divided by sqrt(1-v^2), and so v = sin t is the obvious sub to deal with the denominator.

Zacken essentially did this all in one go, but I freely admit to generally sucking at integration and so I just looked for step-by-step ways of trying to move towards something I could integrate.

Original post by Mad Sentinel

After the sub, 1/r - 1/R becomes

\frac{1}{uR} - \frac{1}{R} = \frac{1}{R} \left(\frac{1}{u} - 1 \right)

and then since R is constant you can take it out of the integral.

Reply 11

Mad Sentinel

Original post by DFranklin

I'm kind of surprised given what I posted that you can't see it's a series of logical steps.

No need for insults buddy!

Anyway, I worked it all through with the substitution that you guys suggested and managed to get the right answer. I haven't done this stuff for a few years so as I said, I'm very rusty.

Reply 12

DFranklin

Original post by Mad Sentinel

No need for insults buddy!I made no insult.

But to my mind, if I suggest 3 successive substitutions and give reasons for what I expect to accomplish by each of the first two, it should be fairly obvious that there is a logical process going on.