What we are trying to find is:
\[ \int \operatorname{arccosec}(x) \ dx \]
If \(y = \operatorname{arccosec}(x)\), then \(\operatorname{cosec}(y) = x\). So:
\[ \frac{dx}{dy} = - \operatorname{cosec}(y) \cot(y) \]
This means:
\[ \int \operatorname{arccosec}(x) \ dx = \int y (- \operatorname{cosec}(y) \cot(y) \ dy) \]
To evaluate this, we can use integration by parts:
\[ \int u \ dv = uv - \int v \ du\]
Let's define \(u\) and \(v\):
\[\begin{gathered} u = y,\ du= dy \\ v = \operatorname{cosec}(y) ,\ dv = \operatorname{cosec}(y) \cot(y) dy \end{gathered}\]
This means:
\[ \int y (- \operatorname{cosec}(y) \cot(y) \ dy) = y \operatorname{cosec}(y) - \int \operatorname{cosec}(y) \ dy\]
The integral of \( \operatorname{cosec}(y)\) is derived here:
\[ \int y (- \operatorname{cosec}(y) \cot(y) \ dy) = y \operatorname{cosec}(y) + \ln | \operatorname{cosec}(y) + \cot(y)| +C\]
Since \(y = \operatorname{arccosec}(x)\):
\[ \int \operatorname{arccosec}(x) \ dx = \operatorname{arccosec}(x) x + \ln |x + \cot(y)| +C\]
Using the Pythagorean theorem, we can say:
\[ \begin{align} \cot^2(y) &= \operatorname{cosec}^2(y) -1 \\ \cot(y) &= ±\sqrt{x^2 -1} \end{align} \]
Therefore:
\[\int \operatorname{arccosec}(x) \ dx = \operatorname{arccosec}(x) x + \ln |x ±\sqrt{x^2 -1}| +C\]
The range of \(y=\operatorname{arccosec}(x)\) is \([-\frac{\pi}{2},0) \cup (0,\frac{\pi}{2}]\). In a unit circle, \(\cot(y)\) and \(x=\operatorname{cosec}(y)\) are either both positive (when \(y \in (0,\frac{\pi}{2}]\)) or both negative (when \(y \in [-\frac{\pi}{2},0) \)).
If \(x\), is negative, then \(\cot(y)\) is also negative:
\[\begin{align} \int \operatorname{arccosec}(x) \ dx &= \operatorname{arccosec}(x) x + \ln (-|x| -|\sqrt{x^2 -1}|) +C \\ &= \operatorname{arccosec}(x) x +\ln (|x| + |\sqrt{x^2 -1}|) +C \end{align}\]
If \(x\), is positive, then \(\cot(y)\) is also positive:
\[ \int \operatorname{arccosec}(x) \ dx = \operatorname{arccosec}(x) x + \ln (|x| + |\sqrt{x^2 -1}|) +C\]
So for the entire domain:
\[ \int \operatorname{arccosec}(x) \ dx = \operatorname{arccosec}(x) x + \ln (|x| + |\sqrt{x^2 -1}|) +C\]