Primitive of Reciprocal of Root of x squared minus a squared/Logarithm Form

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Theorem

$\displaystyle \int \frac {\d x} {\sqrt {x^2 - a^2} } = \map \ln {x + \sqrt {x^2 - a^2} } + C$


Proof 1

\(\displaystyle \int \frac {\d x} {\sqrt {x^2 - a^2} }\) \(=\) \(\displaystyle \cosh^{-1} {\frac x a} + C'\) Primitive of Reciprocal of $\sqrt {x^2 - a^2}$: $\cosh^{-1}$ form
\(\displaystyle \) \(=\) \(\displaystyle \map \ln {\frac x a + \sqrt {\paren {\frac x a}^2 - 1} } + C'\) Definition of Real Inverse Hyperbolic Cosine
\(\displaystyle \) \(=\) \(\displaystyle \map \ln {\frac x a + \sqrt {\frac {x^2 - a^2} {a^2} } } + C'\) rearrangement
\(\displaystyle \) \(=\) \(\displaystyle \map \ln {\frac {x + \sqrt {x^2 - a^2} } a} + C'\) rearrangement
\(\displaystyle \) \(=\) \(\displaystyle \map \ln {x + \sqrt {x^2 - a^2} } - \ln a + C'\) Difference of Logarithms
\(\displaystyle \) \(=\) \(\displaystyle \map \ln {x + \sqrt {x^2 - a^2} } + C\) putting $C = -\ln a + C'$

$\blacksquare$


Proof 2

With a view to using Integration by Substitution:

$\dfrac x a =: \sec \theta: \theta \in \openint 0 {\dfrac \pi 2} \cup \openint {\dfrac \pi 2} \pi$

From Difference of Squares of Secant and Tangent:

$\dfrac {x^2} {a^2} - 1 = \sec^2 \theta - 1 = \tan^2 \theta$

From Shape of Secant Function, this substitution is valid for all $x \in \R \setminus \openint {-1} 1$.

\(\displaystyle x\) \(=\) \(\displaystyle a \sec \theta\)
\(\displaystyle 1\) \(=\) \(\displaystyle a \sec \theta \tan \theta \frac {\rd \theta} {\d x}\) differentiating both sides WRT $x$, Derivative of Secant Function, Chain Rule
\(\displaystyle \int \frac 1 {\sqrt{x^2 - a^2} } \rd x\) \(=\) \(\displaystyle \int \frac {a \sec \theta \tan \theta} {\sqrt{a^2 \sec^2 \theta - a^2} } \frac {\d \theta} {\d x} \rd x\) from above
\(\displaystyle \) \(=\) \(\displaystyle \frac a a \int \frac {\sec \theta \tan \theta} {\sqrt{\sec^2 \theta - 1} } \rd \theta\) Integration by Substitution
\(\displaystyle \) \(=\) \(\displaystyle \int \sec \theta \frac {\tan \theta} {\size {\tan \theta} } \rd \theta\) Difference of Squares of Secant and Tangent


Let $\theta \in \openint 0 {\dfrac \pi 2}$.

Then:

\(\displaystyle \int \sec \theta \rd \theta\) \(=\) \(\displaystyle \ln \size {\sec \theta + \tan \theta} + C\) Primitive of Secant Function
\(\displaystyle \) \(=\) \(\displaystyle \ln \size {\frac x a + \sqrt {\frac {x^2} {a^2} - 1} } + C\)
\(\displaystyle \) \(=\) \(\displaystyle \size {\ln \size {\frac x a + \sqrt {\frac {x^2} {a^2} - 1} } } + C\) Definition of Absolute Value


Let $\theta \in \openint {\dfrac \pi 2} \pi$.

Then:

\(\displaystyle \int -\sec \theta \rd \theta\) \(=\) \(\displaystyle -\ln \size {\sec \theta + \tan \theta} + C\) Primitive of Secant Function

We have that $\dfrac x a < -1$ for $\theta \in \openint {\dfrac \pi 2} \pi$.

But note that $\sqrt {x^2 - a^2}$ is smaller in magnitude than $x$ for $\dfrac x a < -1$.

This puts $x + \sqrt {x^2 - a^2}$ in the domain of $\ln$ such that:

$\ln \size {\dfrac x a + \sqrt {\dfrac {x^2} {a^2} - 1} } < 0$

So:

\(\displaystyle \int \frac {\d x} {\sqrt{x^2 - a^2} }\) \(=\) \(\displaystyle \size {\ln \size {\frac x a + \sqrt {\frac {x^2} {a^2} - 1} } } + C\) Definition of Absolute Value
\(\displaystyle \) \(=\) \(\displaystyle \size {\ln \size {\frac x a + \sqrt {\frac {x^2 - a^2} {a^2} } } } + C\) rearrangement
\(\displaystyle \) \(=\) \(\displaystyle \size {\ln \size {\frac {x + \sqrt {x^2 - a^2} } a} } + C\) rearrangement
\(\displaystyle \) \(=\) \(\displaystyle \size {\ln \size {x + \sqrt {x^2 - a^2} - \ln a} } + C\) Difference of Logarithms
\(\displaystyle \) \(=\) \(\displaystyle \size {\ln \size {x + \sqrt {x^2 - a^2} } } + C\) subsuming $-\ln a$ into constant




Also see


Sources