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 | |||||||||||
\(\ds \) | \(=\) | \(\ds \int \frac {\rd x} {\sqrt{a^2} \sqrt {1 - \paren {\frac x a}^2} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 a \int \frac {\rd x} {\sqrt {1 - \paren {\frac x a}^2} }\) |
- $\cos \theta = \dfrac x a \iff x = a \cos \theta$
for $\theta \in \openint 0 \pi$.
From Real Cosine Function is Bounded and Shape of Cosine Function, this substitution is valid for all $x / a \in \openint {-1} 1$.
\(\ds a^2\) | \(>\) | \(\, \ds x^2 \, \) | \(\ds \) | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds 1\) | \(>\) | \(\, \ds \frac {x^2} {a^2} \, \) | \(\ds \) | 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
- 1944: R.P. Gillespie: Integration (2nd ed.) ... (previous) ... (next): Chapter $\text {II}$: Integration of Elementary Functions: $\S 7$. Standard Integrals: $12$.
- 1974: Murray R. Spiegel: Theory and Problems of Advanced Calculus (SI ed.) ... (previous) ... (next): Chapter $5$. Integrals: Integrals of Special Functions: $25$