Primitive of x by Power of Root of x squared minus a squared
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Theorem
- $\ds \int x \paren {\sqrt {x^2 - a^2} }^n \rd x = \dfrac {\paren {\sqrt {x^2 - a^2} }^{n + 2} } {n + 2} + C$
for $n \ne -2$.
Proof
Let:
\(\ds z\) | \(=\) | \(\ds x^2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d z} {\d x}\) | \(=\) | \(\ds 2 x\) | Derivative of Power | ||||||||||
\(\text {(1)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds \frac {\d z} 2\) | \(=\) | \(\ds x \rd x\) |
Thus:
\(\ds \int x \paren {\sqrt {x^2 - a^2} }^n \rd x\) | \(=\) | \(\ds \int \paren {\sqrt {z - a^2} }^n \dfrac {\d z} 2\) | Integration by Substitution from $(1)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 2 \int \paren {\sqrt {z - a^2} }^n \rd z\) | Primitive of Constant Multiple of Function | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 2 \paren {\frac {2 \paren {\sqrt {z - a^2} }^{n + 2} } {n + 2} } + C\) | Primitive of $\paren {\sqrt {z - a^2} }^n$ for $n \ne -2$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\paren {\sqrt {x^2 - a^2} }^{n + 2} } {n + 2} + C\) | substituting $x^2$ for $z$ and simplifying |
$\blacksquare$
Also see
For $n = -2$, use Primitive of $\dfrac x {x^2 - a^2}$.
Sources
- 1968: George B. Thomas, Jr.: Calculus and Analytic Geometry (4th ed.) ... (previous) ... (next): Front endpapers: A Brief Table of Integrals: $40$.