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 | |||||||||||
\(\ds \) | \(=\) | \(\ds \int \frac {\d x} {\sqrt {a^2} \sqrt {1 - \paren {\frac x a}^2} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 a \int \frac {\d x} {\sqrt {1 - \paren {\frac x a}^2} }\) |
- $\sin \theta = \dfrac x a \iff x = a \sin \theta$
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
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- For a video presentation of the contents of this page, visit the Khan Academy.