Primitive of Reciprocal of a squared minus x squared/Inverse Hyperbolic Tangent Form
Jump to navigation
Jump to search
Theorem
- $\ds \int \frac {\d x} {a^2 - x^2} = \frac 1 a \tanh^{-1} \frac x a + C$
where $\size x < a$.
Proof
Let $\size x < a$.
Let:
\(\ds u\) | \(=\) | \(\ds \tanh^{-1} {\frac x a}\) | Definition of Real Inverse Hyperbolic Tangent, which is defined where $\size {\dfrac x a} < 1$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds a \tanh u\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d x} {\d u}\) | \(=\) | \(\ds a \sech^2 u\) | Derivative of Hyperbolic Tangent | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \int \frac {\rd x} {a^2 - x^2}\) | \(=\) | \(\ds \int \frac {a \sech^2 u} {a^2 - a^2 \tanh^2 u} \rd u\) | Integration by Substitution | ||||||||||
\(\ds \) | \(=\) | \(\ds \frac a {a^2} \int \frac {\sech^2 u} {1 - \tanh^2 u} \rd u\) | Primitive of Constant Multiple of Function | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 a \int \frac {\sech^2 u} {\sech^2 u} \rd u\) | Sum of Squares of Hyperbolic Secant and Tangent | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 a \int \rd u\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 a u + C\) | Integral of Constant | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 a \tanh^{-1} \frac x a + C\) | Definition of $u$ |
Also see
- Primitive of $\dfrac 1 {x^2 - a^2}$: $\coth^{-1}$ form for the case $\size x > a > 0$
Sources
- 1944: R.P. Gillespie: Integration (2nd ed.) ... (previous) ... (next): Chapter $\text {II}$: Integration of Elementary Functions: $\S 7$. Standard Integrals: $16$.
- 1960: Margaret M. Gow: A Course in Pure Mathematics ... (previous) ... (next): Chapter $10$: Integration: $10.4$. Standard integrals: Standard Forms: $\text {(vii)}$
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: General Rules of Integration: $14.41$
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $a^2 - x^2$, $x^2 < a^2$: $14.163$