Primitive of Reciprocal of x by Root of a squared minus x squared/Inverse Hyperbolic Secant Form

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Theorem

For $a > 0$ and $0 < \size x < a$:

$\ds \int \frac {\d x} {x \sqrt {a^2 - x^2} } = -\frac 1 a \sech^{-1} {\frac {\size x} a} + C$


Proof

We note that $\sech^{-1} \dfrac x a$ is defined for $x \in \hointl 0 a$.

Hence we treat the two cases $x > 0$ and $x < 0$ separately.


First let $x > 0$.

Note that $\dfrac 1 {x \sqrt {a^2 - x^2} }$ is not defined at $\pm a$, so we are concerned only about the interval $\openint 0 a$.

Then:

\(\ds u\) \(=\) \(\ds \sech^{-1} {\frac x a}\)
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds a \sech u\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d x} {\d u}\) \(=\) \(\ds -a \sech u \tanh u\) Derivative of Hyperbolic Secant
\(\ds \leadsto \ \ \) \(\ds \int \frac {\d x} {x \sqrt {a^2 - x^2} }\) \(=\) \(\ds \int \frac {-a \sech u \tanh u} {a \sech u \sqrt {a^2 - a^2 \sech^2 u} } \rd u\) Integration by Substitution
\(\ds \) \(=\) \(\ds -\frac a {a^2} \int \frac {\sech u \tanh u} {\sech u \sqrt {1 - \sech^2 u} } \rd u\) Primitive of Constant Multiple of Function
\(\ds \) \(=\) \(\ds -\frac 1 a \int \frac {\sech u \tanh u} {\sech u \tanh u} \rd u\) Sum of Squares of Hyperbolic Secant and Tangent
\(\ds \) \(=\) \(\ds -\frac 1 a \int 1 \rd u\)
\(\ds \) \(=\) \(\ds -\frac 1 a u + C\) Integral of Constant
\(\ds \) \(=\) \(\ds -\frac 1 a \sech^{-1} {\frac x a} + C\) Definition of $u$

$\Box$


Now let $x < 0$.

Let $z = -x$.

Then:

\(\ds z\) \(=\) \(\ds -x\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d z} {\d x}\) \(=\) \(\ds -1\) Derivative of Constant Multiple
\(\ds \leadsto \ \ \) \(\ds \int \frac {\d x} {x \sqrt {a^2 - x^2} }\) \(=\) \(\ds \int \frac {-\d z} {\paren {-z} \sqrt {a^2 - z^2} }\) Integration by Substitution
\(\ds \) \(=\) \(\ds \int \frac {\d z} {z \sqrt {a^2 - z^2} }\) Primitive of Constant Multiple of Function
\(\ds \) \(=\) \(\ds -\frac 1 a \sech^{-1} {\frac z a} + C\) from above: $z \in \hointl 0 a$
\(\ds \) \(=\) \(\ds -\frac 1 a \sech^{-1} {\frac {-x} a} + C\) Definition of $z$

$\Box$


So when $x > 0$:

$\ds \int \frac {\d x} {x \sqrt {a^2 - x^2} } = -\frac 1 a \sech^{-1} {\frac x a} + C$

and when $x < 0$:

$\ds \int \frac {\d x} {x \sqrt {a^2 - x^2} } = -\frac 1 a \sech^{-1} {\frac {-x} a} + C$

It follows that for $\size x \in \openint 0 a$:

$\ds \int \frac {\d x} {x \sqrt {a^2 - x^2} } = -\frac 1 a \sech^{-1} {\frac {\size x} a} + C$

by definition of absolute value.

$\blacksquare$


Also see

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