Primitive of Power of Root of a squared minus x squared

Theorem

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

for $n \ne -1$.

Proof

Let:

\(\ds x\)

\(=\)

\(\ds a \sin \theta\)

\(\text {(1)}: \quad\)

\(\ds \leadsto \ \ \)

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

\(=\)

\(\ds a \cos \theta\)

Derivative of Sine Function

Also:

\(\ds x\)

\(=\)

\(\ds a \sin \theta\)

\(\ds \leadsto \ \ \)

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

\(=\)

\(\ds \sqrt {a^2 \paren {1 - \sin^2 \theta} }\)

\(\ds \)

\(=\)

\(\ds \sqrt {a^2 \cos^2 \theta}\)

Sum of Squares of Sine and Cosine

\(\text {(2)}: \quad\)

\(\ds \)

\(=\)

\(\ds a \cos \theta\)

Thus:

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

\(=\)

\(\ds \int \paren {a \cos \theta}^n \cdot a \cos \theta \rd \theta\)

Integration by Substitution from $(1)$ and $(2)$

\(\ds \)

\(=\)

\(\ds a^{n + 1} \int \cos^{n + 1} \theta \rd \theta\)

Primitive of Constant Multiple of Function

\(\ds \)

\(=\)

\(\ds a^{n + 1} \paren {\frac {\sin \theta \cos^n \theta} {n + 1} - \frac n {n + 1} \int \cos^{n - 1} \theta \rd \theta}\)

Primitive of $\cos^{n + 1} \theta$ for $n \ne -1$

\(\ds \)

\(=\)

\(\ds \frac {a \sin \theta \cdot a^n \cos^n \theta} {n + 1} - a^2 \frac n {n + 1} \int a^{n - 2} \cos^{n - 2} \theta \cdot a \cos \theta \rd \theta\)

rearranging

\(\ds \)

\(=\)

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

substituting for $\sinh \theta$ and $\cosh \theta$

$\blacksquare$

Also see

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