avewright/proofwiki-math huggingface dataset:
Theorem: Primitive of Arccosecant of x over a

:<nowiki>$\ds \int \arccsc \frac x a \rd x = \begin {cases}
x \arccsc \dfrac x a + a \map \ln {x + \sqrt {x^2 - a^2} } + C & : 0 < \arccsc \dfrac x a < \dfrac \pi 2 \\
x \arccsc \dfrac x a - a \map \ln {x + \sqrt {x^2 - a^2} } + C & : -\dfrac \pi 2 < \arccsc \dfrac x a < 0 \\
\end {cases}$
</nowiki>

Proof:
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
{{begin-eqn}}
{{eqn | l = u
      | r = \arccsc \frac x a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = <nowiki>\begin{cases} \dfrac {-a} {x \sqrt {x^2 - a^2} } & : 0 < \arccsc \dfrac x a < \dfrac \pi 2 \\
\dfrac a {x \sqrt {x^2 - a^2} } & : -\dfrac \pi 2 < \arccsc \dfrac x a < 0 \\
\end{cases}</nowiki>
      | c = Derivative of $\arccsc \dfrac x a$
}}
{{end-eqn}}
and let:
{{begin-eqn}}
{{eqn | l = \frac {\d v} {\d x}
      | r = 1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = x
      | c = Primitive of Constant
}}
{{end-eqn}}
First let $\arccsc \dfrac x a$ be in the interval $\openint 0 {\dfrac \pi 2}$.
Then:
{{begin-eqn}}
{{eqn | l = \int \arccsc \frac x a \rd x
      | r = x \arccsc \frac x a - \int x \paren {\frac {-a} {x \sqrt {x^2 - a^2} } } \rd x + C
      | c = Integration by Parts
}}
{{eqn | r = x \arccsc \frac x a + a \int \frac {\d x} {\sqrt {x^2 - a^2} } + C
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = x \arccsc \frac x a + a \map \ln {x + \sqrt {x^2 - a^2} } + C
      | c = Primitive of $\dfrac 1 {\sqrt {x^2 - a^2} }$
}}
{{end-eqn}}
Similarly, let $\arccsc \dfrac x a$ be in the interval $\openint {-\dfrac \pi 2} 0$.
Then:
{{begin-eqn}}
{{eqn | l = \int \arccsc \frac x a \rd x
      | r = x \arccsc \frac x a - \int x \paren {\frac a {x \sqrt {x^2 - a^2} } } \rd x + C
      | c = Integration by Parts
}}
{{eqn | r = x \arccsc \frac x a - a \int \frac {\d x} {\sqrt {x^2 - a^2} } + C
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = x \arccsc \frac x a - a \map \ln {x + \sqrt {x^2 - a^2} } + C
      | c = Primitive of $\dfrac 1 {\sqrt {x^2 - a^2} }$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x by Arcsecant of x over a

:$\ds \int x \arcsec \frac x a \rd x = \begin{cases}
\dfrac {x^2} 2 \arcsec \dfrac x a - \dfrac {a \sqrt {x^2 - a^2} } 2 + C & : 0 < \arcsec \dfrac x a < \dfrac \pi 2 \\
\dfrac {x^2} 2 \arcsec \dfrac x a + \dfrac {a \sqrt {x^2 - a^2} } 2 + C & : \dfrac \pi 2 < \arcsec \dfrac x a < \pi \\
\end{cases}$

Proof:
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
{{begin-eqn}}
{{eqn | l = u
      | r = \arcsec \frac x a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = \begin {cases} \dfrac a {x \sqrt {x^2 - a^2} } & : 0 < \arcsec \dfrac x a < \dfrac \pi 2 \\
\dfrac {-a} {x \sqrt {x^2 - a^2} } & : \dfrac \pi 2 < \arcsec \dfrac x a < \pi \\
\end {cases}
      | c = Derivative of $\arcsec \dfrac x a$
}}
{{end-eqn}}
and let:
{{begin-eqn}}
{{eqn | l = \frac {\d v} {\d x}
      | r = x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = \frac {x^2} 2
      | c = Primitive of Power
}}
{{end-eqn}}
First let $\arcsec \dfrac x a$ be in the interval $\openint 0 {\dfrac \pi 2}$.
Then:
{{begin-eqn}}
{{eqn | l = \int x \arcsec \frac x a \rd x
      | r = \frac {x^2} 2 \arcsec \frac x a - \int \frac {x^2} 2 \paren {\frac a {x \sqrt {x^2 - a^2} } } \rd x + C
      | c = Integration by Parts
}}
{{eqn | r = \frac {x^2} 2 \arcsec \frac x a - \frac a 2 \int \frac {x \rd x} {\sqrt {x^2 - a^2} } + C
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = \frac {x^2} 2 \arcsec \frac x a - \frac a 2 \sqrt {x^2 - a^2} + C
      | c = Primitive of $\dfrac x {\sqrt {x^2 - a^2} }$
}}
{{eqn | r = \frac {x^2} 2 \arcsec \frac x a - \frac {a \sqrt {x^2 - a^2} } 2 + C
      | c = simplifying
}}
{{end-eqn}}
Similarly, let $\arcsec \dfrac x a$ be in the interval $\openint {\dfrac \pi 2} \pi$.
Then:
{{begin-eqn}}
{{eqn | l = \int x \arcsec \frac x a \rd x
      | r = \frac {x^2} 2 \arcsec \frac x a - \int \frac {x^2} 2 \paren {\frac {-a} {x \sqrt {x^2 - a^2} } } \rd x + C
      | c = Integration by Parts
}}
{{eqn | r = \frac {x^2} 2 \arcsec \frac x a + \frac a 2 \int \frac {x \rd x} {\sqrt {x^2 - a^2} } + C
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = \frac {x^2} 2 \arcsec \frac x a + \frac a 2 \sqrt {x^2 - a^2} + C
      | c = Primitive of $\dfrac x {\sqrt {x^2 - a^2} }$
}}
{{eqn | r = \frac {x^2} 2 \arcsec \frac x a + \frac {a \sqrt {x^2 - a^2} } 2 + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x by Arccosecant of x over a

:<nowiki>$\ds \int x \arccsc \frac x a \rd x = \begin{cases}
\dfrac {x^2} 2 \arccsc \dfrac x a + \dfrac {a \sqrt {x^2 - a^2} } 2 + C & : 0 < \arccsc \dfrac x a < \dfrac \pi 2 \\
\dfrac {x^2} 2 \arccsc \dfrac x a - \dfrac {a \sqrt {x^2 - a^2} } 2 + C & : -\dfrac \pi 2 < \arccsc \dfrac x a < 0 \\
\end{cases}$</nowiki>

Proof:
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
{{begin-eqn}}
{{eqn | l = u
      | r = \arccsc \frac x a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = <nowiki>\begin {cases} \dfrac {-a} {x \sqrt {x^2 - a^2} } & : 0 < \arccsc \dfrac x a < \dfrac \pi 2 \\
\dfrac a {x \sqrt {x^2 - a^2} } & : -\dfrac \pi 2 < \arccsc \dfrac x a < 0 \\
\end {cases}</nowiki>
      | c = Derivative of $\arccsc \dfrac x a$
}}
{{end-eqn}}
and let:
{{begin-eqn}}
{{eqn | l = \frac {\d v} {\d x}
      | r = x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = \frac {x^2} 2
      | c = Primitive of Power
}}
{{end-eqn}}
First let $\arccsc \dfrac x a$ be in the interval $\openint 0 {\dfrac \pi 2}$.
Then:
{{begin-eqn}}
{{eqn | l = \int x \arccsc \frac x a \rd x
      | r = \frac {x^2} 2 \arccsc \frac x a - \int \frac {x^2} 2 \paren {\frac {-a} {x \sqrt {x^2 - a^2} } } \rd x + C
      | c = Integration by Parts
}}
{{eqn | r = \frac {x^2} 2 \arccsc \frac x a + \frac a 2 \int \frac {x \rd x} {\sqrt {x^2 - a^2} } + C
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = \frac {x^2} 2 \arccsc \frac x a + \frac a 2 \sqrt {x^2 - a^2} + C
      | c = Primitive of $\dfrac x {\sqrt {x^2 - a^2} }$
}}
{{eqn | r = \frac {x^2} 2 \arccsc \frac x a + \frac {a \sqrt{x^2 - a^2} } 2 + C
      | c = simplifying
}}
{{end-eqn}}
Similarly, let $\arccsc \dfrac x a$ be in the interval $\openint {-\dfrac \pi 2} 0$.
Then:
{{begin-eqn}}
{{eqn | l = \int x \arccsc \frac x a \rd x
      | r = \frac {x^2} 2 \arccsc \frac x a - \int \frac {x^2} 2 \paren {\frac {-a} {x \sqrt {x^2 - a^2} } } \rd x + C
      | c = Integration by Parts
}}
{{eqn | r = \frac {x^2} 2 \arccsc \frac x a - \frac a 2 \int \frac {x \rd x} {\sqrt {x^2 - a^2} } + C
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = \frac {x^2} 2 \arccsc \frac x a - \frac a 2 \sqrt {x^2 - a^2} + C
      | c = Primitive of $\dfrac x {\sqrt {x^2 - a^2} }$
}}
{{eqn | r = \frac {x^2} 2 \arccsc \frac x a - \frac {a \sqrt{x^2 - a^2} } 2 + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x squared by Arcsecant of x over a

:$\ds \int x^2 \arcsec \frac x a \rd x = \begin{cases}
\dfrac {x^3} 3 \arcsec \dfrac x a - \dfrac {a x \sqrt {x^2 - a^2} } 6 - \dfrac {a^3} 6 \map \ln {x + \sqrt {x^2 - a^2} } + C & : 0 < \arcsec \dfrac x a < \dfrac \pi 2 \\
\dfrac {x^3} 3 \arcsec \dfrac x a + \dfrac {a x \sqrt {x^2 - a^2} } 6 + \dfrac {a^3} 6 \map \ln {x + \sqrt {x^2 - a^2} } + C & : \dfrac \pi 2 < \arcsec \dfrac x a < \pi \\
\end{cases}$

Proof:
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
{{begin-eqn}}
{{eqn | l = u
      | r = \arcsec \frac x a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = \begin{cases} \dfrac a {x \sqrt {x^2 - a^2} } & : 0 < \arcsec \dfrac x a < \dfrac \pi 2 \\
\dfrac {-a} {x \sqrt {x^2 - a^2} } & : \dfrac \pi 2 < \arcsec \dfrac x a < \pi \\
\end{cases}
      | c = Derivative of $\arcsec \dfrac x a$
}}
{{end-eqn}}
and let:
{{begin-eqn}}
{{eqn | l = \frac {\d v} {\d x}
      | r = x^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = \frac {x^3} 3
      | c = Primitive of Power
}}
{{end-eqn}}
First let $\arcsec \dfrac x a$ be in the interval $\openint 0 {\dfrac \pi 2}$.
Then:
{{begin-eqn}}
{{eqn | l = \int x^2 \arcsec \frac x a \rd x
      | r = \frac {x^3} 3 \arcsec \frac x a - \int \frac {x^3} 3 \paren {\frac a {x \sqrt {x^2 - a^2} } } \rd x + C
      | c = Integration by Parts
}}
{{eqn | r = \frac {x^3} 3 \arcsec \frac x a - \frac a 3 \int \frac {x^2 \rd x} {\sqrt {x^2 - a^2} } + C
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = \frac {x^3} 3 \arcsec \frac x a - \frac a 3 \paren {\frac {x \sqrt {x^2 - a^2} } 2 + \frac {a^2} 2 \map \ln {x + \sqrt {x^2 - a^2} } } + C
      | c = Primitive of $\dfrac {x^2} {\sqrt {x^2 - a^2} }$
}}
{{eqn | r = \frac {x^3} 3 \arcsec \frac x a - \frac {a x \sqrt{x^2 - a^2} } 6 - \frac {a^3} 6 \map \ln {x + \sqrt {x^2 - a^2} } + C
      | c = simplifying
}}
{{end-eqn}}
Similarly, let $\arcsec \dfrac x a$ be in the interval $\openint {\dfrac \pi 2} \pi$.
Then:
{{begin-eqn}}
{{eqn | l = \int x^2 \arcsec \frac x a \rd x
      | r = \frac {x^3} 3 \arcsec \frac x a - \int \frac {x^3} 3 \paren {\frac {-a} {x \sqrt {x^2 - a^2} } } \rd x + C
      | c = Integration by Parts
}}
{{eqn | r = \frac {x^3} 3 \arcsec \frac x a + \frac a 3 \int \frac {x^2 \rd x} {\sqrt {x^2 - a^2} } + C
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = \frac {x^3} 3 \arcsec \frac x a + \frac a 3 \paren {\frac {x \sqrt {x^2 - a^2} } 2 + \frac {a^2} 2 \map \ln {x + \sqrt {x^2 - a^2} } } + C
      | c = Primitive of $\dfrac {x^2} {\sqrt {x^2 - a^2} }$
}}
{{eqn | r = \frac {x^3} 3 \arcsec \frac x a + \frac {a x \sqrt{x^2 - a^2} } 6 + \frac {a^3} 6 \map \ln {x + \sqrt {x^2 - a^2} } + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x squared by Arccosecant of x over a

:$\ds \int x^2 \arccsc \frac x a \rd x = \begin {cases}
\dfrac {x^3} 3 \arccsc \dfrac x a - \dfrac {a x \sqrt{x^2 - a^2} } 6 - \dfrac {a^3} 6 \map \ln {x + \sqrt {x^2 - a^2} } + C & : 0 < \arccsc \dfrac x a < \dfrac \pi 2 \\
\dfrac {x^3} 3 \arccsc \dfrac x a + \dfrac {a x \sqrt{x^2 - a^2} } 6 + \dfrac {a^3} 6 \map \ln {x + \sqrt {x^2 - a^2} } + C & : -\dfrac \pi 2 < \arccsc \dfrac x a < 0 \\
\end {cases}$

Proof:
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
{{begin-eqn}}
{{eqn | l = u
      | r = \arccsc \frac x a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = \begin {cases} \dfrac {-a} {x \sqrt {x^2 - a^2} } & : 0 < \arccsc \dfrac x a < \dfrac \pi 2 \\
\dfrac a {x \sqrt {x^2 - a^2} } & : -\dfrac \pi 2 < \arccsc \dfrac x a < 0 \\
\end {cases}
      | c = Derivative of $\arccsc \dfrac x a$
}}
{{end-eqn}}
and let:
{{begin-eqn}}
{{eqn | l = \frac {\d v} {\d x}
      | r = x^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = \frac {x^3} 3
      | c = Primitive of Power
}}
{{end-eqn}}
First let $\arccsc \dfrac x a$ be in the interval $\openint 0 {\dfrac \pi 2}$.
Then:
{{begin-eqn}}
{{eqn | l = \int x^2 \arccsc \frac x a \rd x
      | r = \frac {x^3} 3 \arccsc \frac x a - \int \frac {x^3} 3 \paren {\frac {-a} {x \sqrt {x^2 - a^2} } } \rd x + C
      | c = Integration by Parts
}}
{{eqn | r = \frac {x^3} 3 \arccsc \frac x a + \frac a 3 \int \frac {x^2 \rd x} {\sqrt {x^2 - a^2} } + C
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = \frac {x^3} 3 \arccsc \frac x a + \frac a 3 \paren {\frac {x \sqrt {x^2 - a^2} } 2 + \frac {a^2} 2 \map \ln {x + \sqrt {x^2 - a^2} } } + C
      | c = Primitive of $\dfrac {x^2} {\sqrt {x^2 - a^2} }$
}}
{{eqn | r = \frac {x^3} 3 \arccsc \frac x a + \frac {a x \sqrt{x^2 - a^2} } 6 + \frac {a^3} 6 \map \ln {x + \sqrt {x^2 - a^2} } + C
      | c = simplifying
}}
{{end-eqn}}
Similarly, let $\arccsc \dfrac x a$ be in the interval $\openint {-\dfrac \pi 2} 0$.
Then:
{{begin-eqn}}
{{eqn | l = \int x^2 \arccsc \frac x a \rd x
      | r = \frac {x^3} 3 \arccsc \frac x a - \int \frac {x^3} 3 \paren {\frac a {x \sqrt {x^2 - a^2} } } \rd x + C
      | c = Integration by Parts
}}
{{eqn | r = \frac {x^3} 3 \arccsc \frac x a - \frac a 3 \int \frac {x^2 \rd x} {\sqrt {x^2 - a^2} } + C
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = \frac {x^3} 3 \arccsc \frac x a - \frac a 3 \paren {\frac {x \sqrt {x^2 - a^2} } 2 + \frac {a^2} 2 \map \ln {x + \sqrt {x^2 - a^2} } } + C
      | c = Primitive of $\dfrac {x^2} {\sqrt {x^2 - a^2} }$
}}
{{eqn | r = \frac {x^3} 3 \arccsc \frac x a - \frac {a x \sqrt{x^2 - a^2} } 6 - \frac {a^3} 6 \map \ln {x + \sqrt {x^2 - a^2} } + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Arcsecant of x over a over x

<onlyinclude>
{{begin-eqn}}
{{eqn | l = \int \dfrac 1 x \arcsec \frac x a \rd x
      | r = \frac \pi 2 \ln \size x + \sum_{n \mathop \ge 0} \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2 \paren {2 n + 1}^2} \paren {\frac a x}^{2 n + 1} + C
      | c = 
}}
{{eqn | r = \frac \pi 2 \ln \size x + \frac a x + \frac 1 {2 \times 3 \times 3} \paren {\frac a x}^3 + \frac {1 \times 3} {2 \times 4 \times 5 \times 5} \paren {\frac a x}^5 + \frac {1 \times 3 \times 5} {2 \times 4 \times 6 \times 7 \times 7} \paren {\frac a x}^7 + \cdots + C
      | c = 
}}
{{end-eqn}}

Proof:
{{begin-eqn}}
{{eqn | l = \arcsec \frac x a
      | r = \frac \pi 2 - \sum_{n \mathop = 0}^\infty \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2 \paren {2 n + 1} } \paren {\frac a x}^{2 n + 1}
      | c = Power Series Expansion for Real Arcsecant Function
}}
{{eqn | ll= \leadsto
      | l = \frac 1 x \arcsec \frac x a
      | r = \frac \pi {2 x} - \sum_{n \mathop = 0}^\infty \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2 \paren {2 n + 1} } a^{2 n + 1} \paren {\frac 1 x}^{2 n + 2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \int \frac 1 x \arcsec \frac x a \rd x
      | r = \int \paren {\frac \pi {2 x} - \sum_{n \mathop = 0}^\infty \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2 \paren {2 n + 1} } a^{2 n + 1} \paren {\frac 1 x}^{2 n + 2} } \rd x
      | c = 
}}
{{eqn | r = \frac \pi 2 \int \frac {\d x} x - \sum_{n \mathop = 0}^\infty \paren {\int \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2 \paren {2 n + 1} } a^{2 n + 1} \paren {\frac 1 x}^{2 n + 2} \rd x}
      | c = Fubini's Theorem
}}
{{eqn | r = \frac \pi 2 \ln \size x + \sum_{n \mathop = 0}^\infty {\paren {-\int \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2 \paren {2 n + 1} } a^{2 n + 1} \paren {\frac 1 x}^{2 n + 2} \rd x} } + C
      | c = Primitive of Reciprocal
}}
{{eqn | r = \frac \pi 2 \ln \size x + \sum_{n \mathop = 0}^\infty \frac {-\paren {2 n}!} {2^{2 n} \paren {n!}^2 \paren {2 n + 1} \paren {-\paren {2 n + 1} } } a^{2 n + 1} \paren {\frac 1 x}^{2 n + 1}  + C
      | c = Primitive of Power
}}
{{eqn | r = \frac \pi 2 \ln \size x + \sum_{n \mathop \ge 0} \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2 \paren {2 n + 1}^2} \paren {\frac a x}^{2 n + 1} + C
      | c = rearranging
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Arcsecant of x over a over x squared

:$\ds \int \dfrac 1 {x^2} \arcsec \frac x a \rd x = \begin {cases}
-\dfrac 1 x \arcsec \dfrac x a + \dfrac {\sqrt {x^2 - a^2} } {a x} + C & : 0 < \arcsec \dfrac x a < \dfrac \pi 2 \\
-\dfrac 1 x \arcsec \dfrac x a - \dfrac {\sqrt {x^2 - a^2} } {a x} + C & : \dfrac \pi 2 < \arcsec \dfrac x a < \pi \\
\end {cases}$

Proof:
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
{{begin-eqn}}
{{eqn | l = u
      | r = \arcsec \frac x a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = \begin {cases} \dfrac a {x \sqrt {x^2 - a^2} } & : 0 < \arcsec \dfrac x a < \dfrac \pi 2 \\
\dfrac {-a} {x \sqrt {x^2 - a^2} } & : \dfrac \pi 2 < \arcsec \dfrac x a < \pi \\
\end {cases}
      | c = Derivative of $\arcsec \dfrac x a$
}}
{{end-eqn}}
and let:
{{begin-eqn}}
{{eqn | l = \frac {\d v} {\d x}
      | r = \frac 1 {x^2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = \frac {-1} x
      | c = Primitive of Power
}}
{{end-eqn}}
First let $\arcsec \dfrac x a$ be in the interval $\openint 0 {\dfrac \pi 2}$.
Then:
{{begin-eqn}}
{{eqn | l = \int \dfrac 1 {x^2} \arcsec \frac x a
      | r = \frac {-1} x \arcsec \frac x a - \int \frac {-1} x \paren {\frac a {x \sqrt {x^2 - a^2} } } \rd x + C
      | c = Integration by Parts
}}
{{eqn | r = -\dfrac 1 x \arcsec \dfrac x a + a \int \frac {\d x} {x \sqrt {x^2 - a^2} } + C
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = -\dfrac 1 x \arcsec \dfrac x a + a \paren {\frac {\sqrt {x^2 - a^2} } {a^2 x} } + C
      | c = Primitive of $\dfrac 1 {x^2 \sqrt {x^2 - a^2} }$
}}
{{eqn | r = -\dfrac 1 x \arcsec \dfrac x a + \frac {\sqrt {x^2 - a^2} } {a x} + C
      | c = simplifying
}}
{{end-eqn}}
Similarly, let $\arcsec \dfrac x a$ be in the interval $\openint {\dfrac \pi 2} \pi$.
Then:
{{begin-eqn}}
{{eqn | l = \int \dfrac 1 {x^2} \arcsec \frac x a
      | r = \frac {-1} x \arcsec \frac x a - \int \frac {-1} x \paren {\frac {-a} {x \sqrt {x^2 - a^2} } } \rd x + C
      | c = Integration by Parts
}}
{{eqn | r = -\dfrac 1 x \arcsec \dfrac x a - a \int \frac {\d x} {x \sqrt {x^2 - a^2} } + C
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = -\dfrac 1 x \arcsec \dfrac x a - a \paren {\frac {\sqrt {x^2 - a^2} } {a^2 x} } + C
      | c = Primitive of $\dfrac 1 {x^2 \sqrt {x^2 - a^2} }$
}}
{{eqn | r = -\dfrac 1 x \arcsec \dfrac x a - \frac {\sqrt {x^2 - a^2} } {a x} + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Arccosecant of x over a over x squared

:$\ds \int \dfrac 1 {x^2} \arccsc \frac x a \rd x = \begin{cases}
-\dfrac 1 x \arccsc \dfrac x a - \dfrac {\sqrt{x^2 - a^2} } {a x} + C & : 0 < \arccsc \dfrac x a < \dfrac \pi 2 \\
-\dfrac 1 x \arccsc \dfrac x a + \dfrac {\sqrt{x^2 - a^2} } {a x} + C & : -\dfrac \pi 2 < \arccsc \dfrac x a < 0 \\
\end{cases}$

Proof:
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
{{begin-eqn}}
{{eqn | l = u
      | r = \arccsc \frac x a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = \begin {cases} \dfrac {-a} {x \sqrt {x^2 - a^2} } & : 0 < \arccsc \dfrac x a < \dfrac \pi 2 \\
\dfrac a {x \sqrt {x^2 - a^2} } & : -\dfrac \pi 2 < \arccsc \dfrac x a < 0 \\
\end {cases}
      | c = Derivative of $\arccsc \dfrac x a$
}}
{{end-eqn}}
and let:
{{begin-eqn}}
{{eqn | l = \frac {\d v} {\d x}
      | r = \frac 1 {x^2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = \frac {-1} x
      | c = Primitive of Power
}}
{{end-eqn}}
First let $\arccsc \dfrac x a$ be in the interval $\openint 0 {\dfrac \pi 2}$.
Then:
{{begin-eqn}}
{{eqn | l = \int \dfrac 1 {x^2} \arccsc \frac x a
      | r = \frac {-1} x \arccsc \frac x a - \int \frac {-1} x \paren {\frac {-a} {x \sqrt {x^2 - a^2} } } \rd x + C
      | c = Integration by Parts
}}
{{eqn | r = -\dfrac 1 x \arccsc \dfrac x a - a \int \frac {\d x} {x \sqrt {x^2 - a^2} } + C
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = -\dfrac 1 x \arccsc \dfrac x a - a \paren {\frac {\sqrt {x^2 - a^2} } {a^2 x} } + C
      | c = Primitive of $\dfrac 1 {x^2 \sqrt {x^2 - a^2} }$
}}
{{eqn | r = -\dfrac 1 x \arccsc \dfrac x a - \frac {\sqrt{x^2 - a^2} } {a x} + C
      | c = simplifying
}}
{{end-eqn}}
Similarly, let $\arccsc \dfrac x a$ be in the interval $\openint {-\dfrac \pi 2} 0$.
Then:
{{begin-eqn}}
{{eqn | l = \int \dfrac 1 {x^2} \arccsc \frac x a
      | r = \frac {-1} x \arccsc \frac x a - \int \frac {-1} x \paren {\frac a {x \sqrt {x^2 - a^2} } } \rd x + C
      | c = Integration by Parts
}}
{{eqn | r = -\dfrac 1 x \arccsc \dfrac x a + a \int \frac {\d x} {x \sqrt {x^2 - a^2} } + C
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = -\dfrac 1 x \arccsc \dfrac x a + a \paren {\frac {\sqrt {x^2 - a^2} } {a^2 x} } + C
      | c = Primitive of $\dfrac 1 {x^2 \sqrt {x^2 - a^2} }$
}}
{{eqn | r = -\dfrac 1 x \arccsc \dfrac x a + \frac {\sqrt {x^2 - a^2} } {a x} + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Power of x by Arcsine of x over a

:$\ds \int x^m \arcsin \frac x a \rd x = \frac {x^{m + 1} } {m + 1} \arcsin \frac x a - \frac 1 {m + 1} \int \frac {x^{m + 1} \rd x} {\sqrt {a^2 - x^2} }$

Proof:
Recall:
{{begin-eqn}}
{{eqn | n = 1
      | l = \int x^m \arcsin x \rd x
      | r = \frac {x^{m + 1} } {m + 1} \arcsin x - \frac 1 {m + 1} \int \frac {x^{m + 1} \rd x} {\sqrt {1 - x^2} }
      | c = Primitive of $x^m \arcsin x$
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int x^m \arcsin \frac x a \rd x
      | r = \int a^m \paren {\dfrac x a}^m \arcsin \frac x a \rd x
      | c = manipulating into appropriate form
}}
{{eqn | r = a^m \int \paren {\dfrac x a}^m \arcsin \frac x a \rd x
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = a^m \paren {\dfrac 1 {1 / a} \paren {\frac 1 {m + 1} \paren {\dfrac x a}^{m + 1} \arcsin \frac x a - \frac 1 {m + 1} \int \paren {\dfrac x a}^{m + 1} \frac {\d x} {\sqrt {1 - \paren {\dfrac x a}^2} } } }
      | c = Primitive of Function of Constant Multiple, from $(1)$
}}
{{eqn | r = \frac {x^{m + 1} } {m + 1} \arcsin \frac x a - \frac 1 {m + 1} \int \frac {x^{m + 1} \rd x} {\sqrt {a^2 - x^2} } 
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Power of x by Arcsine of x over a

:$\ds \int x^m \arcsin \frac x a \rd x = \frac {x^{m + 1} } {m + 1} \arcsin \frac x a - \frac 1 {m + 1} \int \frac {x^{m + 1} \rd x} {\sqrt {a^2 - x^2} }$

Proof:
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
{{begin-eqn}}
{{eqn | l = u
      | r = \arcsin \frac x a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = \frac 1 {\sqrt {a^2 - x^2} }
      | c = Derivative of $\arcsin \dfrac x a$
}}
{{end-eqn}}
and let:
{{begin-eqn}}
{{eqn | l = \frac {\d v} {\d x}
      | r = x^m
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = \frac {x^{m + 1} } {m + 1}
      | c = Primitive of Power
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int x^m \arcsin \frac x a \rd x
      | r = \frac {x^{m + 1} } {m + 1} \arcsin \frac x a - \int \frac {x^{m + 1} } {m + 1} \paren {\frac 1 {\sqrt {a^2 - x^2} } } \rd x
      | c = Integration by Parts
}}
{{eqn | r = \frac {x^{m + 1} } {m + 1} \arcsin \frac x a - \frac 1 {m + 1} \int \frac {x^{m + 1} \rd x} {\sqrt {a^2 - x^2} }
      | c = Primitive of Constant Multiple of Function
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Power of x by Arccosine of x over a

:$\ds \int x^m \arccos \frac x a \rd x = \frac {x^{m + 1} } {m + 1} \arccos \frac x a + \frac 1 {m + 1} \int \frac {x^{m + 1} \rd x} {\sqrt {a^2 - x^2} }$

Proof:
Recall:
{{begin-eqn}}
{{eqn | n = 1
      | l = \int x^m \arccos x \rd x
      | r = \frac {x^{m + 1} } {m + 1} \arccos x + \frac 1 {m + 1} \int \frac {x^{m + 1} \rd x} {\sqrt {1 - x^2} }
      | c = Primitive of $x^m \arccos x$
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int x^m \arccos \frac x a \rd x
      | r = \int a^m \paren {\dfrac x a}^m \arccos \frac x a \rd x
      | c = manipulating into appropriate form
}}
{{eqn | r = a^m \int \paren {\dfrac x a}^m \arccos \frac x a \rd x
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = a^m \paren {\dfrac 1 {1 / a} \paren {\frac 1 {m + 1} \paren {\dfrac x a}^{m + 1} \arccos \frac x a + \frac 1 {m + 1} \int \paren {\dfrac x a}^{m + 1} \frac {\d x} {\sqrt {1 - \paren {\dfrac x a}^2} } } }
      | c = Primitive of Function of Constant Multiple, from $(1)$
}}
{{eqn | r = \frac {x^{m + 1} } {m + 1} \arccos \frac x a + \frac 1 {m + 1} \int \frac {x^{m + 1} \rd x} {\sqrt {a^2 - x^2} } 
      | c = simplifying
}}
{{end-eqn}}
{{qed}}