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

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Theorem

$\ds \int x^2 \sqrt {x^2 - a^2} \rd x = \frac {x \paren {\sqrt {x^2 - a^2} }^3} 4 + \frac {a^2 x \sqrt {x^2 - a^2} } 8 - \frac {a^4} 8 \ln \size {x + \sqrt {x^2 - a^2} } + C$

for $\size x \ge a$.


Proof

We have that $\sqrt {x^2 - a^2}$ is defined only when $x^2 \ge a^2$, that is, either:

$x \ge a$

or:

$x \le -a$

where it is assumed that $a > 0$.


First let $x \ge a$.

Let:

\(\ds z\) \(=\) \(\ds x^2\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d z} {\d x}\) \(=\) \(\ds 2 x\) Power Rule for Derivatives
\(\ds \leadsto \ \ \) \(\ds \int x^2 \sqrt {x^2 - a^2} \rd x\) \(=\) \(\ds \int \frac {z \sqrt {z - a^2} \rd z} {2 \sqrt z}\) Integration by Substitution
\(\ds \) \(=\) \(\ds \frac 1 2 \int \sqrt z \sqrt {z - a^2} \rd z\) Primitive of Constant Multiple of Function
\(\ds \) \(=\) \(\ds \frac 1 2 \paren {\frac {2 z - a^2} 4 \sqrt z \sqrt {z - a^2} - \frac {a^4} 8 \int \frac {\d z} {\sqrt z \sqrt {z - a^2} } } + C\) Primitive of $\sqrt {\paren {a x + b} \paren {p x + q} }$
\(\ds \) \(=\) \(\ds \frac {2 z - a^2} 8 \sqrt z \sqrt {z - a^2} - \frac {a^4} {16} \paren {2 \map \ln {\sqrt {z - a^2} + \sqrt z} } + C\) Primitive of $\dfrac 1 {\sqrt {\paren {a x + b} \paren {p x + q} } }$
\(\ds \) \(=\) \(\ds \frac {2 z - 2 a^2} 8 \sqrt z \sqrt {z - a^2} + \frac {a^2} 8 \sqrt z \sqrt {z - a^2} - \frac {a^4} 8 \ln \paren {\sqrt {z - a^2} + \sqrt z} + C\)
\(\ds \) \(=\) \(\ds \frac {z - a^2} 4 \sqrt z \sqrt {z - a^2} + \frac {a^2} 8 \sqrt z \sqrt {z - a^2} - \frac {a^4} 8 \ln \paren {\sqrt {z - a^2} + \sqrt z} + C\)
\(\ds \) \(=\) \(\ds \frac {x \paren {\sqrt {x^2 - a^2} }^3} 4 + \frac {a^2 x \sqrt {x^2 - a^2} } 8 - \frac {a^4} 8 \ln \paren {x + \sqrt {x^2 - a^2} } + C\) substituting for $z$
\(\ds \) \(=\) \(\ds \frac {x \paren {\sqrt {x^2 - a^2} }^3} 4 + \frac {a^2 x \sqrt {x^2 - a^2} } 8 - \frac {a^4} 8 \ln \size {x + \sqrt {x^2 - a^2} } + C\) as $x + \sqrt {x^2 - a^2} > 0$


Now suppose $x \le -a$.

Let $z = -x$.

Then:

$\d x = -\d z$

and we then have:


\(\ds \int x^2 \sqrt {x^2 - a^2} \rd x\) \(=\) \(\ds -\int \paren {-z}^2 \sqrt {\paren {-z}^2 - a^2} \rd z\) Integration by Substitution
\(\ds \) \(=\) \(\ds -\int z^2 \sqrt {z^2 - a^2} \rd z\) simplifying
\(\ds \) \(=\) \(\ds -\paren {\frac {z \paren {\sqrt {z^2 - a^2} }^3} 4 + \frac {a^2 z \sqrt {z^2 - a^2} } 8 - \frac {a^4} 8 \map \ln {z + \sqrt {z^2 - a^2} } } + C\) from above
\(\ds \) \(=\) \(\ds -\frac {z \paren {\sqrt {z^2 - a^2} }^3} 4 - \frac {a^2 z \sqrt {z^2 - a^2} } 8 + \frac {a^4} 8 \map \ln {z + \sqrt {z^2 - a^2} } + C\) simplifying
\(\ds \) \(=\) \(\ds -\frac {z \paren {\sqrt {z^2 - a^2} }^3} 4 - \frac {a^2 z \sqrt {z^2 - a^2} } 8 - \frac {a^4} 8 \paren {\map \ln {z - \sqrt {z^2 - a^2} } - \map \ln {a^2} } + C\) Negative of $\map \ln {z + \sqrt {z^2 - a^2} }$
\(\ds \) \(=\) \(\ds -\frac {z \paren {\sqrt {z^2 - a^2} }^3} 4 - \frac {a^2 z \sqrt {z^2 - a^2} } 8 - \frac {a^4} 8 \paren {\map \ln {z - \sqrt {z^2 - a^2} } } + C\) subsuming $\dfrac {a^4 \map \ln {-a^2} } 8$ into constant
\(\ds \) \(=\) \(\ds -\frac {\paren {-x} \paren {\sqrt {\paren {-x}^2 - a^2} }^3} 4 - \frac {a^2 \paren {-x} \sqrt {\paren {-x}^2 - a^2} } 8 - \frac {a^4} 8 \paren {\map \ln {\paren {-x} - \sqrt {\paren {-x}^2 - a^2} } } + C\) substituting back for $x$
\(\ds \) \(=\) \(\ds \frac {x \paren {\sqrt {x^2 - a^2} }^3} 4 + \frac {a^2 x \sqrt {x^2 - a^2} } 8 - \frac {a^4} 8 \map \ln {-\paren {x + \sqrt {x^2 - a^2} } } + C\) simplifying
\(\ds \) \(=\) \(\ds \frac {x \paren {\sqrt {x^2 - a^2} }^3} 4 + \frac {a^2 x \sqrt {x^2 - a^2} } 8 - \frac {a^4} 8 \ln \size {x + \sqrt {x^2 - a^2} } + C\) as $-\paren {x + \sqrt {x^2 - a^2} } > 0$: Definition of Absolute Value

$\blacksquare$


Also presented as

This result is also seen presented in the form:

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


Also see


Sources

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