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

From ProofWiki
Jump to navigation Jump to search

Theorem

$\displaystyle \int x^2 \sqrt {x^2 - a^2} \ \mathrm d x = \frac {x \left({\sqrt {x^2 - a^2} }\right)^3} 4 + \frac {a^2 x \sqrt {x^2 - a^2} } 8 - \frac {a^4} 8 \ln \left({x + \sqrt {x^2 - a^2} }\right) + C$


Proof

Let:

\(\displaystyle z\) \(=\) \(\displaystyle x^2\)
\(\displaystyle \implies \ \ \) \(\displaystyle \frac {\mathrm d z} {\mathrm d x}\) \(=\) \(\displaystyle 2 x\) Power Rule for Derivatives
\(\displaystyle \implies \ \ \) \(\displaystyle \int x^2 \sqrt {x^2 - a^2} \ \mathrm d x\) \(=\) \(\displaystyle \int \frac {z \sqrt {z - a^2} \ \mathrm d z} {2 \sqrt z}\) Integration by Substitution
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 2 \int \sqrt z \sqrt {z - a^2} \ \mathrm d z\) Primitive of Constant Multiple of Function
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 2 \left({\frac {2 z - a^2} 4 \sqrt z \sqrt {z - a^2} - \frac {a^4} 8 \int \frac {\mathrm d z} {\sqrt z \sqrt {z - a^2} } }\right) + C\) Primitive of $\sqrt {\left({a x + b}\right) \left({p x + q}\right) }$
\(\displaystyle \) \(=\) \(\displaystyle \frac {2 z - a^2} 8 \sqrt z \sqrt {z - a^2} - \frac {a^4} {16} \left({2 \ln \left({\sqrt {z - a^2} + \sqrt z}\right) }\right) + C\) Primitive of $\dfrac 1 {\sqrt {\left({a x + b}\right) \left({p x + q}\right) } }$
\(\displaystyle \) \(=\) \(\displaystyle \frac {2 z - 2 a^2} 8 \sqrt z \sqrt {z - a^2} + \frac {a^2} 8 \sqrt z \sqrt {z - a^2} - \frac {a^4} 8 \ln \left({\sqrt {z - a^2} + \sqrt z}\right) + C\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {z - a^2} 4 \sqrt z \sqrt {z - a^2} + \frac {a^2} 8 \sqrt z \sqrt {z - a^2} - \frac {a^4} 8 \ln \left({\sqrt {z - a^2} + \sqrt z}\right) + C\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {x \left({\sqrt {x^2 - a^2} }\right)^3} 4 + \frac {a^2 x \sqrt {x^2 - a^2} } 8 - \frac {a^4} 8 \ln \left({x + \sqrt {x^2 - a^2} }\right) + C\) substituting for $z$

$\blacksquare$


Also see


Sources