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

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