Primitive of Reciprocal of Root of a squared minus x squared/Arccosine Form

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Theorem

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

where $a$ is a strictly positive constant and $a^2 > x^2$.


Proof

\(\ds \int \frac 1 {\sqrt {a^2 - x^2} } \rd x\) \(=\) \(\ds \int \frac {\rd x} {\sqrt {a^2 \paren {1 - \frac {x^2} {a^2} } } }\) factor $a^2$ out of the radicand
\(\ds \) \(=\) \(\ds \int \frac {\rd x} {\sqrt{a^2} \sqrt {1 - \paren {\frac x a}^2} }\)
\(\ds \) \(=\) \(\ds \frac 1 a \int \frac {\rd x} {\sqrt {1 - \paren {\frac x a}^2} }\)

Substitute:

$\cos \theta = \dfrac x a \iff x = a \cos \theta$

for $\theta \in \openint 0 \pi$.

From Real Cosine Function is Bounded and Shape of Cosine Function, this substitution is valid for all $x / a \in \openint {-1} 1$.

By hypothesis:

\(\ds a^2\) \(>\) \(\, \ds x^2 \, \) \(\ds \)
\(\ds \leadstoandfrom \ \ \) \(\ds 1\) \(>\) \(\, \ds \frac {x^2} {a^2} \, \) \(\ds \) dividing both terms by $a^2$
\(\ds \leadstoandfrom \ \ \) \(\ds 1\) \(>\) \(\, \ds \paren {\frac x a}^2 \, \) \(\ds \) Powers of Group Elements
\(\ds \leadstoandfrom \ \ \) \(\ds 1\) \(>\) \(\, \ds \size {\frac x a} \, \) \(\ds \) taking the square root of both terms
\(\ds \leadstoandfrom \ \ \) \(\ds -1\) \(<\) \(\, \ds \frac x a \, \) \(\, \ds < \, \) \(\ds 1\) Negative of Absolute Value

so this substitution will not change the domain of the integrand.


Then:

\(\ds x\) \(=\) \(\ds a \cos \theta\) from above
\(\ds \leadsto \ \ \) \(\ds 1\) \(=\) \(\ds -a \sin \theta \frac {\rd \theta} {\rd x}\) differentiating with respect to $x$, Derivative of Cosine Function, Chain Rule for Derivatives
\(\ds \frac 1 a \int \frac 1 {\sqrt {1 - \paren {\frac x a}^2 } } \rd x\) \(=\) \(\ds \frac 1 a \int \frac {-a \sin \theta} {\sqrt {1 - \cos^2 \theta} } \frac {\rd \theta} {\rd x} \rd x\) from above
\(\ds \) \(=\) \(\ds -\frac a a \int \frac {\sin \theta} {\sqrt {1 - \cos^2 \theta} } \rd \theta\) Integration by Substitution
\(\ds \) \(=\) \(\ds -\int \frac {\sin \theta} {\sqrt {\sin^2 \theta} } \rd \theta\) Sum of Squares of Sine and Cosine
\(\ds \) \(=\) \(\ds -\int \frac {\sin \theta} {\size {\sin \theta} } \rd \theta\)

We have defined $\theta$ to be in the open interval $\openint 0 \pi$.

From Sine and Cosine are Periodic on Reals, $\sin \theta > 0$ for the entire interval. Therefore the absolute value is unnecessary, and the integral simplifies to:

\(\ds -\int \rd \theta\) \(=\) \(\ds -\theta + C\)

As $\theta$ was stipulated to be in the open interval $\openint 0 \pi$:

$\cos \theta = \dfrac x a \iff \theta = \arccos \dfrac x a$

The answer in terms of $x$, then, is:

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

$\blacksquare$


Also see


Sources

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