Inverse trigonometric functions

[PFuser1232]

I am familiar with the importance of the following inverse circular/hyperbolic functions:
$\sin^{-1}$, $\cos^{-1}$, $\tan^{-1}$, $\sinh^{-1}$, $\cosh^{-1}$, $\tanh^{-1}$.
However, I don't really get the point of $\csc^{-1}$, $\coth^{-1}$, and so on.
Given any equation of the form $f(x) = a$, where $f$ is the reciprocal of any circular/hyperbolic function, we can always write the equation as $\frac{1}{f(x)} = \frac{1}{a}$ before solving it with the aid of the six aforementioned inverse functions.
For instance, $\csc{x} = 2$, $0 < x < 2\pi$.
$\csc{x} = 2$
$\sin{x} = \frac{1}{2}$
$x = \frac{\pi}{6}, \frac{5\pi}{6}$
Do we use them to avoid cluttered notation in calculus?
For example, for $x \geq 1$:
$$\int \frac{1}{1 - x^2} dx = \coth^{-1}{x} + C = \frac{1}{2} \ln{\frac{x + 1}{x - 1}} + C$$

 

[Mark44]

I am familiar with the importance of the following inverse circular/hyperbolic functions:
$\sin^{-1}$, $\cos^{-1}$, $\tan^{-1}$, $\sinh^{-1}$, $\cosh^{-1}$, $\tanh^{-1}$.
However, I don't really get the point of $\csc^{-1}$, $\coth^{-1}$, and so on.

If the cosecent function is defined, it might be useful in some circumstances to have an inverse. Same for the hyperbolic cotangent.

Given any equation of the form $f(x) = a$, where $f$ is the reciprocal of any circular/hyperbolic function, we can always write the equation as $\frac{1}{f(x)} = \frac{1}{a}$ before solving it with the aid of the six aforementioned inverse functions.
For instance, $\csc{x} = 2$, $0 < x < 2\pi$.
$\csc{x} = 2$
$\sin{x} = \frac{1}{2}$
$x = \frac{\pi}{6}, \frac{5\pi}{6}$
Do we use them to avoid cluttered notation in calculus?
For example, for $x \geq 1$:
$$\int \frac{1}{1 - x^2} dx = \coth^{-1}{x} + C = \frac{1}{2} \ln{\frac{x + 1}{x - 1}} + C$$