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

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Theorem

$\ds \int \frac {\d x} {\sqrt {a^2 - x^2} } = \arcsin \frac x a + C$

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


Corollary 1

$\ds \int \frac {\d x} {\sqrt {1 - x^2} } = \arcsin x + C$


Corollary 2

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


Corollary 3

$\ds \int \frac {\d x} {\sqrt {2 a x - x^2} } = \arcsin \dfrac {x - a} a + C$


Proof

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

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 $x / a \in \openint {-1} 1$.


Then:

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

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


Then:

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

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:

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

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:

$\ds \int \frac {\d x} {\sqrt {a^2 - x^2} } = \arcsin \frac x a + C$

$\blacksquare$


Also see


Sources