Primitive of Arccosecant of x over a over x squared

From ProofWiki
Jump to navigation Jump to search

Theorem

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


Proof

With a view to expressing the primitive in the form:

$\displaystyle \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$

let:

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


and let:

\(\displaystyle \frac {\d v} {\d x}\) \(=\) \(\displaystyle \frac 1 {x^2}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle v\) \(=\) \(\displaystyle \frac {-1} x\) Primitive of Power


First let $\arccsc \dfrac x a$ be in the interval $\openint 0 {\dfrac \pi 2}$.

Then:

\(\displaystyle \int \frac {\arccsc \frac x a} {x^2}\) \(=\) \(\displaystyle \frac {-1} x \arccsc \frac x a - \int \frac {-1} x \paren {\frac {-a} {x \sqrt {x^2 - a^2} } } \rd x + C\) Integration by Parts
\(\displaystyle \) \(=\) \(\displaystyle \frac {-\arccsc \frac x a} x - a \int \frac {\d x} {x \sqrt {x^2 - a^2} } + C\) Primitive of Constant Multiple of Function
\(\displaystyle \) \(=\) \(\displaystyle \frac {-\arccsc \frac x a} x - a \paren {\frac {\sqrt {x^2 - a^2} } {a^2 x} } + C\) Primitive of $\dfrac 1 {x^2 \sqrt {x^2 - a^2} }$
\(\displaystyle \) \(=\) \(\displaystyle \frac {-\arccsc \frac x a} x - \frac {\sqrt{x^2 - a^2} } {a x} + C\) simplifying


Similarly, let $\arccsc \dfrac x a$ be in the interval $\openint {-\dfrac \pi 2} 0$.

Then:

\(\displaystyle \int \frac {\arccsc \frac x a} {x^2}\) \(=\) \(\displaystyle \frac {-1} x \arccsc \frac x a - \int \frac {-1} x \paren {\frac a {x \sqrt {x^2 - a^2} } } \rd x + C\) Integration by Parts
\(\displaystyle \) \(=\) \(\displaystyle \frac {-\arccsc \frac x a} x + a \int \frac {\d x} {x \sqrt {x^2 - a^2} } + C\) Primitive of Constant Multiple of Function
\(\displaystyle \) \(=\) \(\displaystyle \frac {-\arccsc \frac x a} x + a \paren {\frac {\sqrt {x^2 - a^2} } {a^2 x} } + C\) Primitive of $\dfrac 1 {x^2 \sqrt {x^2 - a^2} }$
\(\displaystyle \) \(=\) \(\displaystyle \frac {-\arccsc \frac x a} x + \frac {\sqrt{x^2 - a^2} } {a x} + C\) simplifying

$\blacksquare$


Also see


Sources