Primitive of Reciprocal of x squared by Root of x squared minus a squared/Proof 1
Jump to navigation
Jump to search
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:
| \(\ds x\) | \(=\) | \(\ds a \cosh \theta\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \frac {\d x} {\d \theta}\) | \(=\) | \(\ds a \sinh \theta\) | Derivative of Hyperbolic Cosine |
Then:
| \(\ds x\) | \(=\) | \(\ds a \cosh \theta\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \sqrt {x^2 - a^2}\) | \(=\) | \(\ds \sqrt {a^2 \paren {\cosh^2 \theta - 1} }\) | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sqrt {a^2 \sinh^2 \theta}\) | Difference of Squares of Hyperbolic Cosine and Sine | |||||||||||
| \(\ds \) | \(=\) | \(\ds a \sinh \theta\) |
Hence:
| \(\ds \int \frac {\d x} {x^2 \sqrt {x^2 - a^2} }\) | \(=\) | \(\ds \int \frac {a \sinh \theta \rd \theta} {a^2 \cosh^2 \theta \cdot a \sinh \theta}\) | Integration by Substitution | |||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac 1 {a^2} \int \frac {\rd \theta} {\cosh^2 \theta}\) | simplifying | |||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac 1 {a^2} \tanh \theta + C\) | Primitive of $\dfrac 1 {\cosh^2 \theta}$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac 1 {a^2} \dfrac {a \sinh \theta} {a \cosh \theta} + C\) | Definition 2 of Hyperbolic Tangent | |||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac 1 {a^2} \dfrac {\sqrt {x^2 - a^2} } x + C\) | substituting for $a \sinh \theta$ and $a \cosh \theta$ |
$\blacksquare$