Yet another motion integral

1. Apr 1, 2006

gulsen

$$\int \frac{dx} { \sqrt{ \frac{1}{x} - \frac{1}{b}} }$$
where b is a positive constant.
Using integrator, I know the result:
$$-\sqrt{bx} \sqrt{b-x} + b\sqrt{b} \tan^{-1}( \frac{\sqrt{x}} {\sqrt{b-x}})$$
but how do I calculate it from scratch?

2. Apr 1, 2006

TD

I'd substitute

$$\frac{1}{x} - \frac{1}{b} = y^2 \Leftrightarrow \frac{1}{x} = y^2 + \frac{1}{b} \Leftrightarrow x = \frac{b}{{by^2 + 1}} \Leftrightarrow dx = \frac{{ - b^2 y}}{{\left( {by^2 + 1} \right)^2 }}$$

It gets rid of the square root and returns a rational function in y.

3. Apr 1, 2006

gulsen

After a few conversions, I ended up with this
$$C \int \frac{du}{(u^2 + 1)^2}$$
Which I couldn't solve. I tried switching
$$u = \sinh(t), du = \cosh(t)dt$$
and
$$\int \frac{dt}{\cosh^3(t)}$$
but got stuck again.

4. Apr 1, 2006

5. Apr 1, 2006

gulsen

Got it done. Thanks!

6. Apr 2, 2006

TD

You're welcome!

7. Apr 3, 2006

dextercioby

The way i see it, there's no need for hyperbolic trigonometry.

$$\int \frac{\sqrt{bx}}{\sqrt{b-x}} \ dx$$

becomes under the substitution $$\sqrt{x} =\sqrt{b} \sin t$$

an integral ~ to $\int \sin^{2} t \ dt$.

Daniel.