Primitive of x squared by Root of a squared minus x squared

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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 1

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

$\blacksquare$


Proof 2

\(\ds z\) \(=\) \(\ds x^2\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d z} {\d x}\) \(=\) \(\ds 2 x\) Power Rule for Derivatives
\(\ds \leadsto \ \ \) \(\ds \int x^2 \sqrt {a^2 - x^2} \rd x\) \(=\) \(\ds \int \frac {z \sqrt {a^2 - z} \rd z} {2 \sqrt z}\) Integration by Substitution
\(\ds \) \(=\) \(\ds \frac 1 2 \int \sqrt {z \paren {a^2 - z} } \rd z\) Primitive of Constant Multiple of Function


Recall:

\(\ds \int \sqrt {\paren {a x + b} \paren {p x + q} } \rd x\) \(=\) \(\ds \frac {2 a p x + b p + a q} {4 a p} \sqrt {\paren {a x + b} \paren {p x + q} } - \frac {\paren {b p - a q}^2} {8 a p} \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }\) Primitive of $\sqrt {\paren {a x + b} \paren {p x + q} }$


Then:

\(\ds \frac {2 a p x + b p + a q} {4 a p}\) \(=\) \(\ds \frac {2 \times 1 \times \paren {-1} z + 0 \times \paren {-1} + 1 \times a^2} {4 \times 1 \times \paren {-1} }\) setting $a \gets 1$, $b \gets 0$, $p \gets -1$, $q \gets a^2$, $x \gets z$
\(\ds \) \(=\) \(\ds \frac {-2 z + a^2} {-4}\) simplifying
\(\text {(1)}: \quad\) \(\ds \) \(=\) \(\ds \frac {2 z - a^2} 4\) simplifying further


Then:

\(\ds \frac 1 2 \int \sqrt {z \paren {a^2 - z} } \rd z\) \(=\) \(\ds \frac 1 2 \paren {\frac {2 z - a^2} 4 \sqrt {z \paren {a^2 - z} } - \frac {\paren {0 \times \paren {-1} + 1 \times a^2}^2} {8 \times 1 \times \paren {-1} } \int \frac {\d z} {\sqrt {z \paren {a^2 - z} } } }\) setting $a \gets 1$, $b \gets 0$, $p \gets -1$, $q \gets a^2$, $x \gets z$
\(\ds \) \(=\) \(\ds \frac {2 z - a^2} 8 \sqrt {z \paren {a^2 - z} } + \frac {a^4} {16} \int \frac {\d z} {\sqrt {z \paren {a^2 - z} } }\) simplifying


Recall:

\(\ds \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }\) \(=\) \(\ds \dfrac {-1} {\sqrt {-a p} } \map \arcsin {\dfrac {2 a p x + b p + a q} {a q - b p} } + C\) Primitive of $\dfrac 1 {\sqrt {\paren {a x + b} \paren {p x + q} } }$ for $a > 0, p < 0$


Then:

\(\ds \frac {2 z - a^2} 8 \sqrt {z \paren {a^2 - z} } + \frac {a^4} {16} \int \frac {\d z} {\sqrt {z \paren {a^2 - z} } }\) \(=\) \(\ds \frac {2 z - a^2} 8 \sqrt {z \paren {a^2 - z} } + \frac {a^4} {16} \paren {\dfrac {-1} {\sqrt {-1 \times \paren {-1} } } \arcsin \frac {2 z - a^2} 4} + C\) from $(1)$, and $a \gets 1$, $b \gets 0$, $p \gets -1$, $q \gets a^2$, $x \gets z$
\(\ds \) \(=\) \(\ds \frac {2 z - a^2} 8 \sqrt {z \paren {a^2 - z} } + \frac {a^4} {16} \paren {-\arcsin \frac {2 z - a^2} 4} + C\) simplifying
\(\ds \) \(=\) \(\ds \frac {2 x^2 - a^2} 8 \sqrt {x^2 \paren {a^2 - x^2} } + \frac {a^4} {16} \paren {-\arcsin \frac {2 x^2 - a^2} 4} + C\) substituting $x^2$ back for $z$
\(\ds \) \(=\) \(\ds \frac {2 x^2 - 2 a^2 + a^2} 8 \sqrt {x^2 \paren {a^2 - x^2} } + \frac {a^4} {16} \paren {-\arcsin \frac {2 x^2 - a^2} 4} + C\)
\(\ds \) \(=\) \(\ds -\frac {a^2 - x^2} 4 x \sqrt {a^2 - x^2} + \frac {a^2 x \sqrt {a^2 - x^2} } 8 + \frac {a^4} {16} \paren {-\arcsin \frac {2 x^2 - a^2} 4} + C\) rearranging
\(\ds \) \(=\) \(\ds -\frac {x \paren {\sqrt {a^2 - x^2} }^3} 4 + \frac {a^2 x \sqrt {a^2 - x^2} } 8 + \frac {a^4} {16} \paren {-\arcsin \frac {2 x^2 - a^2} 4} + C\) simplifying


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Also presented as

This result is also seen presented in the form:

$\ds \int x^2 \sqrt {a^2 - x^2} \rd x = \frac {a^4} 8 \arcsin \frac x a - \frac {x \sqrt {a^2 - x^2} \paren {a^2 - 2 x^2} } 8 + C$


Also see


Sources

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