Primitive of Reciprocal of x by Root of x squared minus a squared/Arcsecant Form
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Theorem
- $\ds \int \frac {\d x} {x \sqrt {x^2 - a^2} } = \frac 1 a \arcsec \size {\frac x a} + C$
for $0 < a < \size x$.
Proof
We have that $\sqrt {x^2 - a^2}$ is defined only when $x^2 > a^2$, that is, either:
- $x > a$
or:
- $x < -a$
where it is assumed that $a > 0$.
Consider the arcsecant substitution:
- $u = \arcsec {\dfrac x a}$
which is defined for all $x$ such that $\size {\dfrac x a} \ge 1$.
That is:
- $\size x \ge a$
and it is seen that $u = \arcsec {\dfrac x a}$ is defined over the whole domain of the integrand.
Hence:
\(\ds u\) | \(=\) | \(\ds \arcsec {\frac x a}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds a \sec u\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d x} {\d u}\) | \(=\) | \(\ds a \sec u \tan u\) | Derivative of Secant Function |
Let $x > a$.
\(\ds \int \frac {\d x} {x \sqrt {x^2 - a^2} }\) | \(=\) | \(\ds \int \frac {a \sec u \tan u} {a \sec u \sqrt {a^2 \sec^2 u - a^2} } \rd u\) | Integration by Substitution | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac a {a^2} \int \frac {\sec u \tan u} {\sec u \sqrt {\sec^2 u - 1} } \rd u\) | Primitive of Constant Multiple of Function | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 a \int \frac {\sec u \tan u} {\sec u \tan u} \rd u\) | Difference of Squares of Secant and Tangent | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 a \int 1 \rd u\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 a u + C\) | Integral of Constant | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 a \arcsec {\frac x a} + C\) | Definition of $u$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 a \arcsec {\frac {\size x} a} + C\) | Definition of Absolute Value: $\size x = x$ for $x > 0$ |
Now suppose $x < -a$.
Let $z = -x$.
Then:
- $\d x = -\d z$
and we then have:
\(\ds \int \frac {\d x} {x \sqrt {x^2 - a^2} }\) | \(=\) | \(\ds \int \frac {-\d z} {\paren {-z} \sqrt {\paren {-z}^2 - a^2} }\) | Integration by Substitution | |||||||||||
\(\ds \) | \(=\) | \(\ds \int \frac {\d z} {z \sqrt {z^2 - a^2} }\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 a \arcsec {\frac z a} + C\) | from above | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 a \arcsec {\frac {\paren {-x} } a} + C\) | substituting back for $x$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 a \arcsec {\frac {\size x} a} + C\) | Definition of Absolute Value: $\size x = -x$ for $x < 0$ |
The result follows.
$\blacksquare$
Also see
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: General Rules of Integration: $14.45$
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\sqrt {x^2 - a^2}$: $14.213$
- 1968: George B. Thomas, Jr.: Calculus and Analytic Geometry (4th ed.) ... (previous) ... (next): Front endpapers: A Brief Table of Integrals: $45$.
- 1972: Frank Ayres, Jr. and J.C. Ault: Theory and Problems of Differential and Integral Calculus (SI ed.) ... (previous) ... (next): Chapter $25$: Fundamental Integration Formulas: $20$.