Primitive of Reciprocal of x squared by Root of x squared minus a squared/Proof 2
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Theorem
- $\ds \int \frac {\d x} {x^2 \sqrt {x^2 - a^2} } = \frac {\sqrt {x^2 - a^2} } {a^2 x} + C$
for $\size x > a$.
Proof
Let:
\(\ds z\) | \(=\) | \(\ds x^2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d z} {\d x}\) | \(=\) | \(\ds 2 x\) | Power Rule for Derivatives | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \int \frac {\d x} {x^2 \sqrt {x^2 - a^2} }\) | \(=\) | \(\ds \int \frac {\d z} {2 z \sqrt z \sqrt {z - a^2} }\) | Integration by Substitution | ||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 2 \int \frac {\d z} {z^{3/2} \sqrt {z - a^2} }\) |
Using Primitive of $\dfrac 1 {u^m \sqrt {a u + b} }$:
- $\ds \int \frac {\d u} {u^m \sqrt {a u + b} } = -\frac {\sqrt {a u + b} } {\paren {m - 1} b u^{m - 1} } - \frac {\paren {2 m - 3} a} {\paren {2 m - 2} b} \int \frac {\d u} {u^{m - 1} \sqrt {a u + b} }$
Setting:
\(\ds u\) | \(:=\) | \(\ds z\) | ||||||||||||
\(\ds m\) | \(:=\) | \(\ds \frac 3 2\) | ||||||||||||
\(\ds a\) | \(:=\) | \(\ds 1\) | ||||||||||||
\(\ds b\) | \(:=\) | \(\ds -a^2\) |
\(\ds \frac 1 2 \int \frac {\d z} {z^{3/2} \sqrt {z - a^2} }\) | \(=\) | \(\ds \frac {-\sqrt {z - a^2} } {2 \paren {\paren {\frac 3 2} - 1} \paren {-a^2} z^{\paren {3/2} - 1} } - \frac {2 \paren {\frac 3 2} - 3} {2 \paren {2 \paren {\frac 3 2} - 2} \paren {-a^2} } \int \frac {\d z} {z^{\paren {\frac 3 2} - 1} \sqrt{z - a^2} } + C\) | Primitive of $ \dfrac 1 {x^m \sqrt {a x + b} }$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {-\sqrt {z - a^2} } {-a^2 z^{1/2} } - 0 + C\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\sqrt {x^2 - a^2} } {a^2 x} + C\) | substituting $x$ back for $z$ |
$\blacksquare$