Compute the indefinite integral of a function whose primite is arccos

Question

While I was studying Kepler's first law, I encountered the following indefinite integral: $$\int \frac{\lvert b \rvert}{x^2 \sqrt{-1+\frac{2a}{x}-\frac{b^2}{x^2}}}\,dx.$$ ($a,b\in \Bbb{R}$, $\lvert b\rvert=\sqrt{b^2}$, $a^2-b^2>0$ and may assume $x>0$.) It is claimed that, up to a constant (clearly), $\arccos\Bigl(\frac{b^2-ax}{x\sqrt{a^2-b^2}}\Bigr)$ is the primitive of the integrand. An easy derivation confirms the assertion. But I was not able to find the primitive by myself. Can you help me to understand how such a computation should be handled, please?

Answer 1

I'm not sure if this goes through rigorously (feel free to yell at me in the comments) but this was my process:

First, complete the square in the denominator: $$\sqrt{-1 + \dfrac{2a}{x} - \dfrac{b^2}{x^2}} = \sqrt{\dfrac{a^2}{b^2}-1 - \dfrac{b^2}{x^2} + \dfrac{2a}{x} - \dfrac{a^2}{b^2}} = \sqrt{\dfrac{a^2-b^2}{b^2}-\left(\dfrac{b^2}{x^2}-2\dfrac{b\cdot a}{x \cdot b} + \dfrac{a^2}{b^2}\right)} = \sqrt{\dfrac{a^2-b^2}{b^2}-\left(\dfrac{b^2-ax}{bx}\right)^2}=\dfrac{\sqrt{a^2-b^2}}{|b|}\sqrt{1-\left(\dfrac{b^2-ax}{x\sqrt{a^2-b^2}}\right)^2}$$

Plugging this into our integrand: $$\int \dfrac{|b|}{x^2 \sqrt{-1 + \dfrac{2a}{x} - \dfrac{b^2}{x^2}}} dx = \int\dfrac{|b|^2}{x^2\sqrt{a^2-b^2}}\cdot \dfrac{1}{\sqrt{1-\left(\dfrac{b^2-ax}{x\sqrt{a^2-b^2}}\right)^2}}dx$$

Notice that by the linearity of absolute values, $|b|^2 = |b||b| = |b^2|$, and because $b^2$ is always nonnegative, $|b^2| = b^2$.

Now, introduce the trigonometric substitution: $$\cos{\theta}=\dfrac{b^2-ax}{x\sqrt{a^2-b^2}}$$

Differentiating on both sides: $$-\sin{\theta} d\theta = \dfrac{x\sqrt{a^2-b^2}(-a) - (b^2-ax)\sqrt{a^2-b^2}}{x^2(a^2-b^2)}dx = \dfrac{-b^2}{x^2\sqrt{a^2-b^2}}dx$$

Notice the similarity between this and our integrand's first fraction. Therefore: $$\int\dfrac{b^2}{x^2\sqrt{a^2-b^2}}\cdot \dfrac{1}{\sqrt{1-\left(\dfrac{b^2-ax}{x\sqrt{a^2-b^2}}\right)^2}}dx = \int \dfrac{-1}{\sqrt{1-\cos^2{\theta}}} \cdot -\sin{\theta} d\theta$$

Now, supposing that $\sin\theta > 0$: $$\int \dfrac{-1}{\sqrt{1-\cos^2{\theta}}} \cdot -\sin{\theta} d\theta = \int \dfrac{\sin\theta}{|\sin\theta|} d\theta = \int d\theta = \theta + C$$

By definition, $\cos\theta = \dfrac{b^2-ax}{x\sqrt{a^2-b^2}}$, so $\theta = \arccos\left(\dfrac{b^2-ax}{x\sqrt{a^2-b^2}}\right)$.

$$\int \dfrac{|b|}{x^2 \sqrt{-1 + \dfrac{2a}{x} - \dfrac{b^2}{x^2}}} dx = \arccos\left(\dfrac{b^2-ax}{x\sqrt{a^2-b^2}}\right) + C$$

I believe this works, I am a little unsure about the finer details of assuming $\sin\theta > 0$ and whether the use of $\arccos$ is appropriate, so there may be some details left which need to be filled in, but I would say that this is at least a solid motivation for the given primitive.