Calculating a Motion Integral Without Hyperbolic Trigonometry

[gulsen]

$$\int \frac{dx} { \sqrt{ \frac{1}{x} - \frac{1}{b}} }$$
where b is a positive constant.
Using http://integrals.wolfram.com" , 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?

 

[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.

 

[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.

 

[TD]

Take a look at https://www.physicsforums.com/showpost.php?p=951501&postcount=11" and the rest of that topic

 

[gulsen]

Got it done. Thanks!

 

[TD]

Got it done. Thanks!

You're welcome!

 

[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.