Primitive of Root of a squared minus x squared over x squared

Theorem

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

Proof

Let:

\(\ds z\)

\(=\)

\(\ds x^2\)

\(\ds \leadsto \ \ \)

\(\ds \frac {\d z} {\d x}\)

\(=\)

\(\ds 2 x\)

Power Rule for Derivatives

\(\ds \leadsto \ \ \)

\(\ds \int \frac {\sqrt {a^2 - x^2} } {x^2} \rd x\)

\(=\)

\(\ds \int \frac {\sqrt {a^2 - z} \rd z} {2 z \sqrt z}\)

Integration by Substitution

\(\ds \)

\(=\)

\(\ds \frac 1 2 \int \frac {\sqrt {a^2 - z} \rd z} {z^{3/2} }\)

Primitive of Constant Multiple of Function

\(\ds \)

\(=\)

\(\ds \frac 1 2 \paren {\frac {-\sqrt {a^2 - z} } {\frac 1 2 \sqrt z} - \frac 1 2 \int \frac {\d z} {\sqrt z \sqrt {a^2 - z} } } + C\)

Primitive of $\dfrac {\sqrt {a x + b} } {x^m}$

\(\ds \)

\(=\)

\(\ds \frac {-\sqrt {a^2 - x^2} } x - \frac 1 2 \int \frac {2 x \rd x} {x \sqrt {a^2 - x^2} } + C\)

substituting for $z$

\(\ds \)

\(=\)

\(\ds \frac {-\sqrt {a^2 - x^2} } x - \int \frac {\d x} {\sqrt {a^2 - x^2} } + C\)

simplifying

\(\ds \)

\(=\)

\(\ds \frac {-\sqrt {a^2 - x^2} } x - \arcsin \frac x a + C\)

Primitive of $\dfrac 1 {\sqrt {a^2 - x^2} }$

$\blacksquare$

Also see

• Primitive of $\dfrac {\sqrt {x^2 + a^2} } {x^2}$
• Primitive of $\dfrac {\sqrt {x^2 - a^2} } {x^2}$

Sources

• 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\sqrt {a^2 - x^2}$: $14.249$
• 1968: George B. Thomas, Jr.: Calculus and Analytic Geometry (4th ed.) ... (previous) ... (next): Front endpapers: A Brief Table of Integrals: $32$.