Primitive of x squared by Arccotangent of x over a
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Theorem
- $\ds \int x^2 \arccot \frac x a \rd x = \frac {x^3} 3 \arccot \frac x a + \frac {a x^2} 6 - \frac {a^3} 6 \map \ln {x^2 + a^2} + C$
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:
\(\ds u\) | \(=\) | \(\ds \arccot \frac x a\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d u} {\d x}\) | \(=\) | \(\ds \frac {-a} {x^2 + a^2}\) | Derivative of $\arccot \dfrac x a$ |
and let:
\(\ds \frac {\d v} {\d x}\) | \(=\) | \(\ds x^2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds v\) | \(=\) | \(\ds \frac {x^3} 3\) | Primitive of Power |
Then:
\(\ds \int x^2 \arccot \frac x a \rd x\) | \(=\) | \(\ds \frac {x^3} 3 \arccot \frac x a - \int \frac {x^3} 3 \paren {\frac {-a} {x^2 + a^2} } \rd x + C\) | Integration by Parts | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {x^3} 3 \arccot \frac x a + \frac a 3 \int \frac {x^3 \rd x} {x^2 + a^2} + C\) | Primitive of Constant Multiple of Function | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {x^3} 3 \arccot \frac x a + \frac a 3 \paren {\frac {x^2} 2 - \frac {a^2} 2 \map \ln {x^2 + a^2} } + C\) | Primitive of $\dfrac {x^3} {x^2 + a^2}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {x^3} 3 \arccot \frac x a + \frac {a x^2} 6 - \frac {a^3} 6 \map \ln {x^2 + a^2} + C\) | simplifying |
$\blacksquare$
Also see
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving Inverse Trigonometric Functions: $14.490$