# Primitive of Reciprocal of Root of x squared minus a squared/Logarithm Form/Proof 2

## Theorem

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

## Proof

With a view to using Integration by Substitution:

$\dfrac x a =: \sec \theta: \theta \in \openint 0 {\dfrac \pi 2} \cup \openint {\dfrac \pi 2} \pi$
$\dfrac {x^2} {a^2} - 1 = \sec^2 \theta - 1 = \tan^2 \theta$

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 for Derivatives $\displaystyle \int \frac 1 {\sqrt{x^2 - a^2} } \rd x$ $=$ $\displaystyle \int \frac {a \sec \theta \tan \theta} {\sqrt{a^2 \sec^2 \theta - a^2} } \frac {\d \theta} {\d x} \rd x$ 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