Solving $x = \frac13\sec(a)$ for $a$ (after integrating via trig substitution) For the integral $\dfrac{1}{x\sqrt{9x^2-1}}$, I decided to use $x = \frac13\sec(a)$. The expression simplifies to the integral of $1$, simply becoming $a$.
The issue is making $a$ the subject of $x = \frac13\sec(a)$. By drawing a triangle of hypotenuse $3x$, base length $1$ and height of $(9x^2 -1)^{1/2}$, where $a$ is the angle between $3x$ and $1$. I end up with three expressions for $a$:
$$a = \arctan\left(\sqrt{9x^2-1}\right) \quad\text{or}\quad
\arcsin\left(\dfrac{\sqrt{9x^2-1}}{3x}\right)\quad\text{or}\quad 
\arccos\left(\dfrac{1}{3x}\right)$$
After graphing all three on Desmos, only the first answer was right, however I don’t know why, and is it always the arctan that’s right? What assumption is being made for the sine and cosine variation that makes the integral incorrect?
Thank you.

First of all you should include $C$ as a constant since it is an idefinite integral.
Now you know that the domain of $\arcsin$ and $\arcsin$ differ from that of $\arctan$ which is defined for all values of $x$. I think the problem might be the way you are plotting the solutions. You have to specify which range of $x$ values you are using.
