Solve Maths Problem: Integral Substitution with Change of Limits

[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 $\frac{\pi}{2}$?