Primitive of Reciprocal of Root of a squared minus x squared

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Theorem

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

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


Corollary

$\displaystyle \int_0^x \frac {\d t} {\sqrt{1 - t^2} } = \arcsin x$


Proof

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

Substitute:

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

for $\theta \in \openint {-\dfrac \pi 2} {\dfrac \pi 2}$.

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

By hypothesis:

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

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


Then:

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

We have defined $\theta$ to be in the open interval $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$.

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

\(\displaystyle \int \rd \theta\) \(=\) \(\displaystyle \theta + C\)

As $\theta$ was stipulated to be in the open interval $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$:

$\sin \theta = \dfrac x a \iff \theta = \arcsin \dfrac x a$

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

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

$\blacksquare$


Also see


Sources