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

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