Primitive of Arccosine of x over a over x squared

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Theorem

$\ds \int \frac 1 {x^2} \arccos \frac x a \rd x = -\frac 1 x \arccos \frac x a + \frac 1 a \map \ln {\frac {a + \sqrt {a^2 - x^2} } x} + C$


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:

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


Then:

\(\ds \int \frac 1 {x^2} \arccos \frac x a \rd x\) \(=\) \(\ds \arccos \frac x a \paren {\frac {-1} x} - \int \paren {\frac {-1} x} \paren {\frac {-1} {\sqrt {a^2 - x^2} } } \rd x + C\) Integration by Parts
\(\ds \) \(=\) \(\ds -\frac 1 x \arccos \frac x a - \int \frac {\d x} {x \sqrt {a^2 - x^2} } \rd x + C\) simplifying
\(\ds \) \(=\) \(\ds -\frac 1 x \arccos \frac x a + \frac 1 a \map \ln {\frac {a + \sqrt {a^2 - x^2} } {\size x} } + C\) Primitive of $\dfrac 1 {x \sqrt {a^2 - x^2} }$
\(\ds \) \(=\) \(\ds -\frac 1 x \arccos \frac x a + \frac 1 a \map \ln {\frac {a + \sqrt {a^2 - x^2} } x} + C\) Domain of Real Arccosine is Non-Negative

$\blacksquare$


Also see


Sources

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