Integral Resulting in Arcsine

Theorem

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

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

Proof

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \int \frac 1 {\sqrt{a^2 - x^2} } \ \mathrm dx\)	\(=\)	\(\displaystyle \int \frac {\mathrm dx}{\sqrt{a^2 \left({1 - \frac {x^2}{a^2} }\right)} }\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	factor $a^2$ out of the radicand
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \int \frac {\mathrm dx}{\sqrt{a^2} \sqrt{1 - \left({\frac x a}\right)^2} }\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac 1 a \int \frac {\mathrm dx}{\sqrt{1 - \left({\frac x a}\right)^2} }\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Substitute:

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

for $\theta \in \left({-\dfrac \pi 2 .. \dfrac \pi 2}\right)$.

From Boundedness of Sine and Cosine and Shape of Sine Function, this substitution is valid for all $x/a \in \left(-1..1\right)$.

By hypothesis:

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

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

Then:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle x\)	\(=\)	\(\displaystyle a \sin \theta\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	from above
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle 1\)	\(=\)	\(\displaystyle a \cos \theta \frac {\mathrm d \theta}{\mathrm dx}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	differentiate WRT $x$, Derivative of Sine Function, Chain Rule
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \frac 1 a \int \frac 1 {\sqrt{1 - \left({\frac x a}\right)^2 } } \ \mathrm dx\)	\(=\)	\(\displaystyle \frac 1 a \int \frac {a \cos \theta} {\sqrt {1 - \sin^2 \theta} } \frac{\mathrm d \theta} {\mathrm d x} \ \mathrm d x\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	from above
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac a a \int \frac{\cos \theta} {\sqrt {1 - \sin^2 \theta} } \ \mathrm d \theta\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Integration by Substitution
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \int \frac {\cos \theta} {\sqrt {\cos^2 \theta} } \ \mathrm d \theta\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Sum of Squares of Sine and Cosine
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \ \int \frac {\cos \theta} {\left \vert \cos \theta \right \vert} \ \mathrm d \theta\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

We have defined $\theta$ to be in the open interval $\left({-\dfrac \pi 2 .. \dfrac \pi 2}\right)$.

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:

\(\displaystyle \)

\(\displaystyle \)

\(\displaystyle \)

\(\displaystyle \int \mathrm d \theta\)

\(=\)

\(\displaystyle \theta + C\)

\(\displaystyle \)

\(\displaystyle \)

\(\displaystyle \)

As $\theta$ was stipulated to be in the open interval $\left({-\dfrac \pi 2 .. \dfrac \pi 2}\right)$:

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

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

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

$\blacksquare$

Also see

• Derivative of Arcsine Function

Sources

• For a video presentation of the contents of this page, visit the Khan Academy.