Primitive of x cubed over Root of a squared minus x squared

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Theorem

$\displaystyle \int \frac {x^3 \rd x} {\sqrt {a^2 - x^2} } = \frac {\left({\sqrt {a^2 - x^2} }\right)^3} 3 - a^2 \sqrt {a^2 - x^2} + C$


Proof

With a view to expressing the problem in the form:

$\displaystyle \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$

let:

\(\displaystyle u\) \(=\) \(\displaystyle x^2\)
\(\displaystyle \implies \ \ \) \(\displaystyle \frac {\d u} {\d x}\) \(=\) \(\displaystyle 2 x\) Power Rule for Derivatives


and let:

\(\displaystyle \frac {\d v} {\d x}\) \(=\) \(\displaystyle \frac x {\sqrt {a^2 - x^2} }\)
\(\displaystyle \implies \ \ \) \(\displaystyle v\) \(=\) \(\displaystyle -\sqrt {a^2 - x^2}\) Primitive of $\dfrac x {\sqrt {a^2 - x^2} }$


Then:

\(\displaystyle \int \frac {x^3 \rd x} {\sqrt {a^2 - x^2} }\) \(=\) \(\displaystyle \int x^2 \frac {x \rd x} {\sqrt {a^2 - x^2} }\)
\(\displaystyle \) \(=\) \(\displaystyle -x^2 \sqrt {a^2 - x^2} - \int 2 x \left({-\sqrt {a^2 - x^2} }\right) \rd x\) Integration by Parts
\(\displaystyle \) \(=\) \(\displaystyle -x^2 \sqrt {a^2 - x^2} + 2 \int x \left({\sqrt {a^2 - x^2} }\right) \rd x\) simplification
\(\displaystyle \) \(=\) \(\displaystyle -x^2 \sqrt {a^2 - x^2} + 2 \left({-\frac {\left({\sqrt {a^2 - x^2} }\right)^3} 3 }\right) + C\) Primitive of $x \sqrt {a^2 - x^2}$
\(\displaystyle \) \(=\) \(\displaystyle \left({a^2 - x^2 - a^2}\right) \sqrt {a^2 - x^2} - \frac {2 \left({\sqrt {a^2 - x^2} }\right)^3} 3 + C\)
\(\displaystyle \) \(=\) \(\displaystyle \left({\sqrt {a^2 - x^2} }\right)^3 - a^2 \sqrt {a^2 - x^2} - \frac {2 \left({\sqrt {a^2 - x^2} }\right)^3} 3 + C\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\left({\sqrt {a^2 - x^2} }\right)^3} 3 - a^2 \sqrt {a^2 - x^2} + C\)

$\blacksquare$


Also see


Sources