Simple maths stuff

[tyco05]

This seems so simple, but I can't for the life of me work it out. I've forgotten a lot of maths over the years, so I need a little help.

The question goes:

Using the substitution

u=\frac{r}{R_1}

Show that :

\int_R^{R_1} \sqrt{\frac{R_1}{r}-1} dr = \int_\frac{R}{R_1}^1 \sqrt{\frac{1}{u}-1} du


So, the substitution under the root is easy enough, it's just the changing of the limits that I can't seem to figure out (or perhaps remember).

I got

\frac{du}{dr}=\frac{1}{R_1}

dr={R_1}du

So substituting

\int_R^{R_1} \sqrt{\frac{R_1}{r}-1} dr = R_1\int_R^{R_1}\sqrt{\frac{1}{u}-1} du


I can see that if I divide both limits by the factor outside the integral, R_1, I'll get the right answer, but surely I can't just do that.
Any hints?

Cheers

[e(ho0n3]

The limits are r = R and r = R1. The corresponding limits after the substitution are u = R/R1 and u = 1 (since uR1 = r). The factor of R1 remains though (in your last integral). I don't see how it could disappear, but who knows...

[tyco05]

thanks, I knew it was reasonably trivial, I just couldn't see it.

That factor of R1 is still supposed to be there though right?
Maybe another mistake on this assignment....

I also need help with the next part, which I don't even know where to start on.

Show that in the limit \frac{R}{R_1}\rightarrow0

\int_{\frac{R}{R_1}}^1\sqrt{\frac{1}{u}-1}du \approx \frac{\pi}{2}-\int_0^{\frac{R}{R_1}}\sqrt{\frac{1}{u}}du

I have no bloody idea where to start, it's doing my head in! It can't be that difficult!!
Any hints?

[Tide]

\int _x^1 \sqrt {\frac {1}{u} -1} du = \int _0^1\sqrt {\frac {1}{u} -1} du - \int_0^x \sqrt {\frac {1}{u} -1} du

The first integral is \frac {\pi}{2} and for the second observe that
\sqrt {\frac {1}{u} - 1} = \sqrt {\frac { 1-u}{u}}
and use the binomial expansion on the numerator. I think there's a factor of \frac {1}{2} missing in the last term of your equation.

[tyco05]

How do you just know that the first integral is \frac{\pi}{2}?