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

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Theorem

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

for $x^2 > a^2$.


Proof

When $x = a$ we have that $\sqrt {x^2 - a^2} = 0$ and then $\dfrac 1 {\sqrt {x^2 - a^2} }$ is not defined.

When $\size x < a$ we have that $x^2 - a^2 < 0$ and then $\sqrt {x^2 - a^2}$ is not defined.

Hence the domain needs to be restricted to $\size x > a$, or that is: $\size {\dfrac x a} > 1$.


Let $x > a$.

Let:

\(\ds u\) \(=\) \(\ds \map \arcosh {\frac x a}\)
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds a \cosh u\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d x} {\d u}\) \(=\) \(\ds a \sinh u\) Derivative of Hyperbolic Cosine Function
\(\ds \leadsto \ \ \) \(\ds \int \frac {\d x} {\sqrt {x^2 - a^2} }\) \(=\) \(\ds \int \frac {a \sinh u} {\sqrt {a^2 \cosh^2 u - a^2} } \rd u\) Integration by Substitution
\(\ds \) \(=\) \(\ds \frac a a \int \frac {\sinh u} {\sqrt {\cosh^2 u - 1} } \rd u\) Primitive of Constant Multiple of Function
\(\ds \) \(=\) \(\ds \int \frac {\sinh u} {\sinh u} \rd u\) Difference of Squares of Hyperbolic Cosine and Sine
\(\ds \) \(=\) \(\ds \int 1 \rd u\)
\(\ds \) \(=\) \(\ds u + C\) Integral of Constant
\(\ds \) \(=\) \(\ds \map \arcosh {\frac x a} + C\) Definition of $u$
\(\ds \) \(=\) \(\ds \arcosh \size {\frac x a} + C\) as $\dfrac x a > 0$
\(\ds \) \(=\) \(\ds \dfrac {\size x} x \arcosh \size {\frac x a} + C\) as $x > 0$, so $\dfrac {\size x} x = 1$


Let $x < -a$.

Let $z = -x$.

Hence:

$\dfrac {\d z} {\d x} = -1$

Then:

\(\ds \int \frac {\d x} {\sqrt {x^2 - a^2} }\) \(=\) \(\ds \int \frac {-\d z} {\sqrt {\paren {-z}^2 - a^2} }\) Integration by Substitution
\(\ds \) \(=\) \(\ds -\int \frac {\d z} {\sqrt {z^2 - a^2} }\) Primitive of Constant Multiple of Function
\(\ds \) \(=\) \(\ds -\map \arcosh {\frac z a} + C\) from above
\(\ds \) \(=\) \(\ds -\map \arcosh {\frac {-x} a} + C\) substituting back
\(\ds \) \(=\) \(\ds -\arcosh \size {\frac x a} + C\) as $\dfrac x a < 0$
\(\ds \) \(=\) \(\ds \dfrac {\size x} x \arcosh \size {\frac x a} + C\) as $x < 0$, so $\dfrac {\size x} x = -1$

$\blacksquare$


Also see


Sources

but beware that he glosses over the details of what happens for negative $x$.
who does the same for negative $x$.
who covers positive $x$ only