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

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Theorem

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


Proof

With a view to expressing the problem in the form:

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

let:

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


and let:

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


Then:

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

$\blacksquare$


Also see


Sources

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