Primitive of x by Arccosine of x over a/Proof 2
Jump to navigation
Jump to search
Theorem
- $\ds \int x \arccos \frac x a \rd x = \paren {\frac {x^2} 2 - \frac {a^2} 4} \arccos \frac x a - \frac {x \sqrt {a^2 - x^2} } 4 + C$
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:
\(\ds u\) | \(=\) | \(\ds \arccos \frac x a\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d u} {\d x}\) | \(=\) | \(\ds \frac {-1} {\sqrt {a^2 - x^2} }\) | Derivative of $\arccos \dfrac x a$ |
and let:
\(\ds \frac {\d v} {\d x}\) | \(=\) | \(\ds x\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds v\) | \(=\) | \(\ds \frac {x^2} 2\) | Primitive of Power |
Then:
\(\ds \int x \arccos \frac x a \rd x\) | \(=\) | \(\ds \frac {x^2} 2 \arccos \frac x a - \int \frac {x^2} 2 \paren {\frac {-1} {\sqrt {a^2 - x^2} } } \rd x + C\) | Integration by Parts | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {x^2} 2 \arccos \frac x a + \frac 1 2 \int \frac {x^2 \rd x} {\sqrt {a^2 - x^2} } + C\) | Primitive of Constant Multiple of Function | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {x^2} 2 \arccos \frac x a + \frac 1 2 \paren {\frac {-x \sqrt {a^2 - x^2} } 2 + \frac {a^2} 2 \arcsin \frac x a} + C\) | Primitive of $\dfrac {x^2} {\sqrt {a^2 - x^2} }$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {x^2} 2 \arccos \frac x a - \frac {x \sqrt {a^2 - x^2} } 4 + \frac {a^2} 4 \paren {\frac \pi 2 - \arccos \frac x a} + C\) | Sum of Arcsine and Arccosine | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\frac {x^2} 2 - \frac {a^2} 4} \arccos \frac x a - \frac {x \sqrt {a^2 - x^2} } 4 + \frac {\pi a^2} 8 + C\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\frac {x^2} 2 - \frac {a^2} 4} \arccos \frac x a - \frac {x \sqrt {a^2 - x^2} } 4 + C\) | subsuming $\dfrac {\pi a^2} 8$ into the arbitrary constant |
$\blacksquare$