Primitive of x squared by Arccosecant of x over a
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Theorem
- $\ds \int x^2 \arccsc \frac x a \rd x = \begin {cases}
\dfrac {x^3} 3 \arccsc \dfrac x a - \dfrac {a x \sqrt{x^2 - a^2} } 6 - \dfrac {a^3} 6 \map \ln {x + \sqrt {x^2 - a^2} } + C & : 0 < \arccsc \dfrac x a < \dfrac \pi 2 \\ \dfrac {x^3} 3 \arccsc \dfrac x a + \dfrac {a x \sqrt{x^2 - a^2} } 6 + \dfrac {a^3} 6 \map \ln {x + \sqrt {x^2 - a^2} } + C & : -\dfrac \pi 2 < \arccsc \dfrac x a < 0 \\ \end {cases}$
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:
\(\ds u\) | \(=\) | \(\ds \arccsc \frac x a\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d u} {\d x}\) | \(=\) | \(\ds \begin {cases} \dfrac {-a} {x \sqrt {x^2 - a^2} } & : 0 < \arccsc \dfrac x a < \dfrac \pi 2 \\
\dfrac a {x \sqrt {x^2 - a^2} } & : -\dfrac \pi 2 < \arccsc \dfrac x a < 0 \\ \end {cases}\) |
Derivative of $\arccsc \dfrac x a$ |
and let:
\(\ds \frac {\d v} {\d x}\) | \(=\) | \(\ds x^2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds v\) | \(=\) | \(\ds \frac {x^3} 3\) | Primitive of Power |
First let $\arccsc \dfrac x a$ be in the interval $\openint 0 {\dfrac \pi 2}$.
Then:
\(\ds \int x^2 \arccsc \frac x a \rd x\) | \(=\) | \(\ds \frac {x^3} 3 \arccsc \frac x a - \int \frac {x^3} 3 \paren {\frac {-a} {x \sqrt {x^2 - a^2} } } \rd x + C\) | Integration by Parts | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {x^3} 3 \arccsc \frac x a + \frac a 3 \int \frac {x^2 \rd x} {\sqrt {x^2 - a^2} } + C\) | Primitive of Constant Multiple of Function | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {x^3} 3 \arccsc \frac x a + \frac a 3 \paren {\frac {x \sqrt {x^2 - a^2} } 2 + \frac {a^2} 2 \map \ln {x + \sqrt {x^2 - a^2} } } + C\) | Primitive of $\dfrac {x^2} {\sqrt {x^2 - a^2} }$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {x^3} 3 \arccsc \frac x a + \frac {a x \sqrt{x^2 - a^2} } 6 + \frac {a^3} 6 \map \ln {x + \sqrt {x^2 - a^2} } + C\) | simplifying |
Similarly, let $\arccsc \dfrac x a$ be in the interval $\openint {-\dfrac \pi 2} 0$.
Then:
\(\ds \int x^2 \arccsc \frac x a \rd x\) | \(=\) | \(\ds \frac {x^3} 3 \arccsc \frac x a - \int \frac {x^3} 3 \paren {\frac a {x \sqrt {x^2 - a^2} } } \rd x + C\) | Integration by Parts | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {x^3} 3 \arccsc \frac x a - \frac a 3 \int \frac {x^2 \rd x} {\sqrt {x^2 - a^2} } + C\) | Primitive of Constant Multiple of Function | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {x^3} 3 \arccsc \frac x a - \frac a 3 \paren {\frac {x \sqrt {x^2 - a^2} } 2 + \frac {a^2} 2 \map \ln {x + \sqrt {x^2 - a^2} } } + C\) | Primitive of $\dfrac {x^2} {\sqrt {x^2 - a^2} }$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {x^3} 3 \arccsc \frac x a - \frac {a x \sqrt{x^2 - a^2} } 6 - \frac {a^3} 6 \map \ln {x + \sqrt {x^2 - a^2} } + C\) | simplifying |
$\blacksquare$
Also see
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving Inverse Trigonometric Functions: $14.500$