Given that $\int \frac{1}{x\sqrt{x^2-1}}dx=\arccos(\frac{1}{x})+C$, what is $\int\frac{1}{x\sqrt{x^2-a^2}}dx$?

Question

The following is clear:

$x\sqrt{x^2-a^2}=x\sqrt{a^2}\sqrt{\frac{x^2}{a^2}-1}=ax\sqrt{\frac{x^2}{a^2}-1}= a^2\frac{x}{a}\sqrt{\frac{x^2}{a^2}-1}$.

So I get that $$\int\frac{1}{x\sqrt{x^2-a^2}}dx=\frac{1}{a^2}\int\frac{1}{\frac{x}{a}\sqrt{\frac{x^2}{a^2}-1}}dx=\frac{1}{a^2}\arccos(\frac{a}{x})+C$$

Is this correct?

Answer 1

You are almost there. Consider $x/a =z.$ This gives $dx = a dz$. That means one of a's will be cancelled. Note that I am assuming that the given number $a$ is nonnegative. Anyway, your solution is correct. You just need to adjust the coefficient with $cos^{-1}(x/a)$.

Answer 2

$$ \frac{1}{a^2}\int\frac{1}{\frac{x}{a}\sqrt{\frac{x^2}{a^2}-1}}dx $$ Substitute $u=\frac{x}{a} \Longrightarrow dx = a*du $

$$ \frac{1}{a}\int\frac{1}{u\sqrt{u^{2}-1}}du = \frac{1}{a} \arccos(\frac{1}{u})= \frac{1}{a} \arccos(\frac{a}{x})$$