Primitive of x by Arccosecant of x over a

From ProofWiki
Jump to navigation Jump to search

Theorem

$\displaystyle \int x \operatorname{arccsc} \frac x a \ \mathrm d x = \begin{cases} \displaystyle \frac {x^2} 2 \operatorname{arccsc} \frac x a + \frac {a \sqrt{x^2 - a^2} } 2 + C & : 0 < \operatorname{arccsc} \dfrac x a < \dfrac \pi 2 \\ \displaystyle \frac {x^2} 2 \operatorname{arccsc} \frac x a - \frac {a \sqrt{x^2 - a^2} } 2 + C & : -\dfrac \pi 2 < \operatorname{arccsc} \dfrac x a < 0 \\ \end{cases}$


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:

\(\displaystyle u\) \(=\) \(\displaystyle \operatorname{arccsc} \frac x a\)
\(\displaystyle \implies \ \ \) \(\displaystyle \frac {\mathrm d u} {\mathrm d x}\) \(=\) \(\displaystyle \begin{cases} \dfrac {-a} {x \sqrt {x^2 - a^2} } & : 0 < \operatorname{arccsc} \dfrac x a < \dfrac \pi 2 \\ \dfrac a {x \sqrt {x^2 - a^2} } & : -\dfrac \pi 2 < \operatorname{arccsc} \dfrac x a < 0 \\ \end{cases}\) Derivative of $\operatorname{arccsc} \dfrac x a$


and let:

\(\displaystyle \frac {\mathrm d v} {\mathrm d x}\) \(=\) \(\displaystyle x\)
\(\displaystyle \implies \ \ \) \(\displaystyle v\) \(=\) \(\displaystyle \frac {x^2} 2\) Primitive of Power


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

Then:

\(\displaystyle \int x \operatorname{arccsc} \frac x a \ \mathrm d x\) \(=\) \(\displaystyle \frac {x^2} 2 \operatorname{arccsc} \frac x a - \int \frac {x^2} 2 \left({\frac {-a} {x \sqrt {x^2 - a^2} } }\right) \ \mathrm d x + C\) Integration by Parts
\(\displaystyle \) \(=\) \(\displaystyle \frac {x^2} 2 \operatorname{arccsc} \frac x a + \frac a 2 \int \frac {x \ \mathrm d x} {\sqrt {x^2 - a^2} } + C\) Primitive of Constant Multiple of Function
\(\displaystyle \) \(=\) \(\displaystyle \frac {x^2} 2 \operatorname{arccsc} \frac x a + \frac a 2 \sqrt {x^2 - a^2} + C\) Primitive of $\dfrac x {\sqrt {x^2 - a^2} }$
\(\displaystyle \) \(=\) \(\displaystyle \frac {x^2} 2 \operatorname{arccsc} \frac x a + \frac {a \sqrt{x^2 - a^2} } 2 + C\) simplifying


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

Then:

\(\displaystyle \int x \operatorname{arccsc} \frac x a \ \mathrm d x\) \(=\) \(\displaystyle \frac {x^2} 2 \operatorname{arccsc} \frac x a - \int \frac {x^2} 2 \left({\frac {-a} {x \sqrt {x^2 - a^2} } }\right) \ \mathrm d x + C\) Integration by Parts
\(\displaystyle \) \(=\) \(\displaystyle \frac {x^2} 2 \operatorname{arccsc} \frac x a - \frac a 2 \int \frac {x \ \mathrm d x} {\sqrt {x^2 - a^2} } + C\) Primitive of Constant Multiple of Function
\(\displaystyle \) \(=\) \(\displaystyle \frac {x^2} 2 \operatorname{arccsc} \frac x a - \frac a 2 \sqrt {x^2 - a^2} + C\) Primitive of $\dfrac x {\sqrt {x^2 - a^2} }$
\(\displaystyle \) \(=\) \(\displaystyle \frac {x^2} 2 \operatorname{arccsc} \frac x a - \frac {a \sqrt{x^2 - a^2} } 2 + C\) simplifying

$\blacksquare$


Also see


Sources