Primitive of Power of Root of x squared minus a squared

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Theorem

$\ds \int \paren {\sqrt {x^2 - a^2} }^n \rd x = \dfrac {x \paren {\sqrt {x^2 - a^2} }^n} {n + 1} - \dfrac {n a^2} {n + 1} \int \paren {\sqrt {x^2 - a^2} }^{n - 2} \rd x$

for $n \ne -1$


Proof

Let:

\(\ds x\) \(=\) \(\ds a \cosh \theta\)
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \frac {\d x} {\d \theta}\) \(=\) \(\ds a \sinh \theta\) Derivative of Hyperbolic Cosine


Also:

\(\ds x\) \(=\) \(\ds a \cosh \theta\)
\(\ds \leadsto \ \ \) \(\ds x^2 - a^2\) \(=\) \(\ds a^2 \cosh^2 \theta - a^2\)
\(\ds \) \(=\) \(\ds a^2 \paren {\cosh^2 \theta - 1}\)
\(\ds \) \(=\) \(\ds a^2 \sinh^2 \theta\) Difference of Squares of Hyperbolic Cosine and Sine
\(\text {(2)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \paren {\sqrt {x^2 - a^2} }^n\) \(=\) \(\ds a^n \sinh^n \theta\)


Thus:

\(\ds \int \paren {\sqrt {x^2 - a^2} }^n \rd x\) \(=\) \(\ds \int \paren {\sqrt {x^2 - a^2} }^n \, a \sinh \theta \rd \theta\) Integration by Substitution from $(1)$
\(\ds \) \(=\) \(\ds \int a^{n + 1} \sinh^{n + 1} \theta \rd \theta\) substituting for $\paren {\sqrt {x^2 - a^2} }^n$ from $(2)$
\(\ds \) \(=\) \(\ds a^{n + 1} \int \sinh^{n + 1} \theta \rd \theta\) Primitive of Constant Multiple of Function
\(\ds \) \(=\) \(\ds a^{n + 1} \paren {\frac {\sinh^n \theta \cosh \theta} {n + 1} - \frac n {n + 1} \int \sinh^{n - 1} \theta \rd \theta}\) Primitive of $\sinh^{n + 1} \theta$ for $n \ne -1$
\(\ds \) \(=\) \(\ds \frac {a^n \sinh^n \theta \cdot a \cosh \theta} {n + 1} - a^2 \frac n {n + 1} \int a^{n - 2} \sinh^{n - 2} \theta \cdot a \sinh \theta \rd \theta\) rearranging
\(\ds \) \(=\) \(\ds \frac {x \paren {\sqrt {x^2 - a^2} }^n} {n + 1} - \frac {n a^2} {n + 1} \int \paren {\sqrt {x^2 - a^2} }^{n - 2} \rd x\) substituting for $\sinh \theta$ and $\cosh \theta$

$\blacksquare$


Also see

For $n = -1$, use Primitive of Reciprocal of Root of x squared minus a squared


Sources

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