Integral of $$$\operatorname{acsc}{\left(x \right)}$$$

For the integral $$$\int{\operatorname{acsc}{\left(x \right)} d x}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Let $$$\operatorname{u}=\operatorname{acsc}{\left(x \right)}$$$ and $$$\operatorname{dv}=dx$$$.

Then $$$\operatorname{du}=\left(\operatorname{acsc}{\left(x \right)}\right)^{\prime }dx=- \frac{\left|{x}\right|}{x^{2} \sqrt{x^{2} - 1}} dx$$$ (steps can be seen ») and $$$\operatorname{v}=\int{1 d x}=x$$$ (steps can be seen »).

Therefore,

$${\color{red}{\int{\operatorname{acsc}{\left(x \right)} d x}}}={\color{red}{\left(\operatorname{acsc}{\left(x \right)} \cdot x-\int{x \cdot \left(- \frac{\left|{x}\right|}{x^{2} \sqrt{x^{2} - 1}}\right) d x}\right)}}={\color{red}{\left(x \operatorname{acsc}{\left(x \right)} - \int{\left(- \frac{\left|{x}\right|}{x \sqrt{x^{2} - 1}}\right)d x}\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=-1$$$ and $$$f{\left(x \right)} = \frac{1}{\sqrt{x^{2} - 1}}$$$:

$$x \operatorname{acsc}{\left(x \right)} - {\color{red}{\int{\left(- \frac{\left|{x}\right|}{x \sqrt{x^{2} - 1}}\right)d x}}} = x \operatorname{acsc}{\left(x \right)} - {\color{red}{\left(- \int{\frac{1}{\sqrt{x^{2} - 1}} d x}\right)}}$$

Let $$$x=\cosh{\left(u \right)}$$$.

Then $$$dx=\left(\cosh{\left(u \right)}\right)^{\prime }du = \sinh{\left(u \right)} du$$$ (steps can be seen »).

Also, it follows that $$$u=\operatorname{acosh}{\left(x \right)}$$$.

Integrand becomes

$$$\frac{1}{\sqrt{x^{2} - 1}} = \frac{1}{\sqrt{\cosh^{2}{\left( u \right)} - 1}}$$$

Use the identity $$$\cosh^{2}{\left( u \right)} - 1 = \sinh^{2}{\left( u \right)}$$$:

$$$\frac{1}{\sqrt{\cosh^{2}{\left( u \right)} - 1}}=\frac{1}{\sqrt{\sinh^{2}{\left( u \right)}}}$$$

Assuming that $$$\sinh{\left( u \right)} \ge 0$$$, we obtain the following:

$$$\frac{1}{\sqrt{\sinh^{2}{\left( u \right)}}} = \frac{1}{\sinh{\left( u \right)}}$$$

Integral becomes

$$x \operatorname{acsc}{\left(x \right)} + {\color{red}{\int{\frac{1}{\sqrt{x^{2} - 1}} d x}}} = x \operatorname{acsc}{\left(x \right)} + {\color{red}{\int{1 d u}}}$$

Apply the constant rule $$$\int c\, du = c u$$$ with $$$c=1$$$:

$$x \operatorname{acsc}{\left(x \right)} + {\color{red}{\int{1 d u}}} = x \operatorname{acsc}{\left(x \right)} + {\color{red}{u}}$$

Recall that $$$u=\operatorname{acosh}{\left(x \right)}$$$:

$$x \operatorname{acsc}{\left(x \right)} + {\color{red}{u}} = x \operatorname{acsc}{\left(x \right)} + {\color{red}{\operatorname{acosh}{\left(x \right)}}}$$

Therefore,

$$\int{\operatorname{acsc}{\left(x \right)} d x} = x \operatorname{acsc}{\left(x \right)} + \operatorname{acosh}{\left(x \right)}$$

Add the constant of integration:

$$\int{\operatorname{acsc}{\left(x \right)} d x} = x \operatorname{acsc}{\left(x \right)} + \operatorname{acosh}{\left(x \right)}+C$$

$$$\int \operatorname{acsc}{\left(x \right)}\, dx = \left(x \operatorname{acsc}{\left(x \right)} + \operatorname{acosh}{\left(x \right)}\right) + C$$$A