Primitive of Arccosecant of x over a

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Theorem

$\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}$


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:

\(\ds u\) \(=\) \(\ds \arccsc \frac x a\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d u} {\d x}\) \(=\) \(\ds \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:

\(\ds \frac {\d v} {\d x}\) \(=\) \(\ds 1\)
\(\ds \leadsto \ \ \) \(\ds v\) \(=\) \(\ds x\) Primitive of Constant


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

Then:

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


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

Then:

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

$\blacksquare$


Also see


Sources

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