Primitive of x squared by Root of a squared minus x squared/Proof 1
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Theorem
- $\ds \int x^2 \sqrt {a^2 - x^2} \rd x = -\frac {x \paren {\sqrt {a^2 - x^2} }^3} 4 + \frac {a^2 x \sqrt {a^2 - x^2} } 8 + \frac {a^4} 8 \arcsin \frac x a + C$
Proof
Let us assume that $a > 0$.
Let:
\(\ds x\) | \(=\) | \(\ds a \sin t\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac {\d x} {\d t}\) | \(=\) | \(\ds a \cos t\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds a^2 - x^2\) | \(=\) | \(\ds a^2 - a^2 \sin^2 t\) | |||||||||||
\(\ds \) | \(=\) | \(\ds a^2 \paren {1 - \sin^2 t}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds a^2 \cos^2 t\) | Sum of Squares of Sine and Cosine |
Hence:
\(\ds \int x^2 \sqrt {a^2 - x^2} \rd x\) | \(=\) | \(\ds \int a^2 \sin^2 t \cdot \sqrt {a^2 \cos^2 t} \cdot a \cos t \rd t\) | Integration by Substitution | |||||||||||
\(\ds \) | \(=\) | \(\ds a^4 \int \sin^2 t \cos^2 t \rd t\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds a^4 \paren {\frac t 8 - \frac {\sin 4 t} {32} } + C\) | Primitive of $\sin^2 a t \cos^2 a t$ where $a \gets 1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds a^4 \paren {\frac t 8 - \dfrac {4 \sin t \cos t - 8 \sin^3 t \cos t} {32} } + C\) | Quadruple Angle Formula for Sine | |||||||||||
\(\text {(1)}: \quad\) | \(\ds \) | \(=\) | \(\ds \dfrac {a^4} 8 \paren {t - \sin t \cos t - 2 \sin^3 t \cos t} + C\) | simplifying |
Recall:
\(\ds x\) | \(=\) | \(\ds a \sin t\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds t\) | \(=\) | \(\ds \arcsin \frac x a\) |
Hence:
\(\ds \) | \(\) | \(\ds \frac {a^4} 8 \paren {t - \sin t \cos t - 2 \sin^3 t \cos t} + C\) | from $(1)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {a^4} 8 \paren {\arcsin \frac x a - \paren {\frac x a} \paren {\frac {\sqrt {a^2 - x^2} } a} + 2 \paren {\frac x a}^3 \paren {\frac {\sqrt {a^2 - x^2} } a} } + C\) | substituting $\sin t = \dfrac x a$, $\cos t = \dfrac {\sqrt {a^2 - x^2} } a$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {a^4} 8 \paren {\arcsin \frac x a + \paren {\frac x a} \paren {\frac {\sqrt {a^2 - x^2} } a} \paren {-1 + \frac {2 x^2} {a^2} } } + C\) | Distributive Laws of Arithmetic | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {a^4} 8 \paren {\arcsin \frac x a + \paren {\frac x a} \paren {\frac {\sqrt {a^2 - x^2} } a} \paren {\dfrac {-2 a^2 + 2 x^2 + a^2} {a^2} } } + C\) | manipulating as necessary | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {a^4} 8 \paren {\arcsin \frac x a + \paren {\frac x a} \frac {\sqrt {a^2 - x^2} } a \paren {-2 \frac {a^2 - x^2} {a^2} + 1} } + C\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\frac {x \paren {\sqrt {a^2 - x^2} }^3} 4 + \frac {a^2 x \sqrt {a^2 - x^2} } 8 + \frac {a^4} 8 \arcsin \frac x a + C\) | simplifying and rearranging |
$\blacksquare$
Also see
Sources
- J. W. Perry (https://math.stackexchange.com/users/93144/j-w-perry), Find a primitive of $x^2\sqrt{a^2 - x^2}$, URL (version: 2013-11-22): https://math.stackexchange.com/q/576068