indefinite integral of $1/(x \sqrt{(x^2 - b^2)})$

Question

EDIT2: Ok, looks like there is really a convention I wasn't aware of. I always looked at "-1" as if it was normal power of "-1". But for trigonometric functions it has entirely different meaning. My bad.

I never thought I would ask something like this, but here am I.

I ran into this problem accidentally while solving one phys.chem. exercise.

Here is the expression from the title again $$\frac{1}{x\cdot\sqrt{x^2 - b^2}}$$

So, according to one table from the internet its indefinite integral equals $$\frac{1}{b} \cdot \cos^{-1}\frac{b}{x}$$

according to Maple it is $$-\frac{\arctan{\frac{b}{\sqrt{x^2-b^2}}}}{b}$$ according to Matlab it is $$\frac{\arctan{\frac{\sqrt{x^2-b^2}}{b}}}{b}$$ Ok, I believe the result from Maple = result from Matlab + some constant, I won't even check it. The derivatives of both results are clearly the original expression, that's what matters.

But hey, what about the table integral? Is it a blatant error or what? Ok, let's manually find the derivative of the following expression: $$\frac{1}{b} \cdot \cos^{-1}\frac{b}{x}$$

I got this: $$-\frac{\sin\frac{b}{x}}{\cos^2{\frac{b}{x}}} \cdot \frac{1}{x^2}$$

Matlab returns the same. However Maple gives different results depending on how you express this power of minus one. Proof: LINK TO IMAGE

Just look at the first one, it is our original expression again! This means that our table wasn't that wrong after all? Or was it? Please solve this riddle for me. I can't believe that 1 expression can have 2 completely different indefinite integrals.

EDIT: and now to finish off your sanity behold another wonder of Maple LINK TO IMAGE mind the reverted argument of secant.

Answer 1

$$\int\frac{1}{x\cdot\sqrt{x^2 - b^2}}\ dx$$

Let $x = 1/t$, $\dfrac{dx}{dt} = -\dfrac{1}{t^2}$

$$-\int\frac{t}{\sqrt{(1/t)^2 - b^2}}\ \dfrac{dt}{t^2} = -\int\frac{t^2}{\sqrt{1 - (tb)^2}}\ \dfrac{dt}{t^2} = -\int\frac{1}{\sqrt{1 - (tb)^2}}\ dt =-\dfrac1b(\sin^{-1}(tb))+ C$$

Using $\sin^{-1} x = \pi/2 - \cos^{-1} x $

$$-\dfrac1b(\sin^{-1}(tb))+ C = \dfrac1b\cos^{-1}\left(\dfrac{b}x\right)+ C$$

Answer 2

Let $ \theta =\cos^{-1} (b/x)$ then \begin{eqnarray*} \cos \theta = \frac{b}{x} \\ \sec^2 \theta = \frac{x^2}{b^2} \\ \tan^2 \theta =\sec^2 \theta -1 =\frac{ x^2}{b^2}-1 \\ \tan \theta = \frac{\sqrt{ x^2-b^2}}{b} \\ \theta = \tan^{-1} \left( \frac{\sqrt{ x^2-b^2}}{b} \right). \end{eqnarray*}