Primitive of x squared by Root of a squared minus x squared/Proof 2

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

\(\ds z\)

\(=\)

\(\ds x^2\)

\(\ds \leadsto \ \ \)

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

\(=\)

\(\ds 2 x\)

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 see

• Primitive of $x^2 \sqrt {x^2 + a^2}$
• Primitive of $x^2 \sqrt {x^2 - a^2}$