Primitive of Arccosecant of x over a

Theorem

 * $\displaystyle \int \operatorname{arccsc} \frac x a \ \mathrm d x = \begin{cases}

\displaystyle x \operatorname{arccsc} \frac x a + a \ln \left({x + \sqrt {x^2 - a^2} }\right) + C & : 0 < \operatorname{arccsc} \dfrac x a < \dfrac \pi 2 \\ \displaystyle x \operatorname{arccsc} \frac x a - a \ln \left({x + \sqrt {x^2 - a^2} }\right) + C & : -\dfrac \pi 2 < \operatorname{arccsc} \dfrac x a < 0 \\ \end{cases}$

Proof
With a view to expressing the primitive in the form:
 * $\displaystyle \int u \frac {\mathrm d v}{\mathrm d x} \ \mathrm d x = u v - \int v \frac {\mathrm d u}{\mathrm d x} \ \mathrm d x$

let:

and let:

First let $\operatorname{arccsc} \dfrac x a$ be in the interval $\left({0 \,.\,.\,\dfrac \pi 2}\right)$.

Then:

Similarly, let $\operatorname{arccsc} \dfrac x a$ be in the interval $\left({-\dfrac \pi 2 \,.\,.\, 0}\right)$.

Then:

Also see

 * Primitive of $\arcsin \dfrac x a$


 * Primitive of $\arccos \dfrac x a$


 * Primitive of $\arctan \dfrac x a$


 * Primitive of $\operatorname{arccot} \dfrac x a$


 * Primitive of $\operatorname{arcsec} \dfrac x a$