Integral of One Over Square Root of Binomial

Theorem

$\displaystyle \int \frac 1 {\sqrt{x^2 - a^2}} \ \mathrm dx = \left \vert{\ln \left \vert{\frac x a + \sqrt {\frac {x^2}{a^2} - 1} }\right \vert}\right \vert + C$

where $a$ is a strictly positive constant and $x^2 > a^2$.

Proof

Substitute:

$\dfrac x a = \sec \theta$, $\theta \in (0..\pi/2) \cup (\pi/2..\pi)$

From Corollary to Sum of Squares of Sine and Cosine, we also have:

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

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

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle x\)	\(=\)	\(\displaystyle a\sec \theta\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle 1\)	\(=\)	\(\displaystyle a\sec \theta \tan \theta \frac {\mathrm d \theta}{\mathrm dx}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Differentiate both sides WRT $x$, Derivative of Secant Function, Chain Rule
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \int \frac 1 {\sqrt{x^2 - a^2} } \ \mathrm dx\)	\(=\)	\(\displaystyle \int \frac {a\sec \theta \tan \theta}{\sqrt{a^2\sec^2 \theta - a^2} } \frac {\mathrm d \theta}{\mathrm dx} \ \mathrm dx\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	from above
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac a a \int \frac {\sec \theta \tan \theta}{\sqrt{\sec^2 \theta - 1} } \ \mathrm d \theta\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Integration by Substitution
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \int \sec \theta \frac {\tan \theta}{\left \vert{\tan \theta} \right \vert} \ \mathrm d \theta\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Corollary to Sum of Squares of Sine and Cosine

Suppose $\theta \in (0..\pi/2)$.

Then:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \int \sec \theta \ \mathrm d \theta\)	\(=\)	\(\displaystyle \ln \left \vert {\sec \theta + \tan \theta} \right \vert + C\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Integral of Secant Function
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \ln \left \vert{\frac x a + \sqrt {\frac {x^2}{a^2} - 1} } \right \vert + C\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left \vert {\ln \left \vert{\frac x a + \sqrt {\frac {x^2}{a^2} - 1} } \right \vert} \right \vert + C\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of Absolute Value

Suppose $\theta \in (\pi/2..\pi)$.

Then:

\(\displaystyle \)

\(\displaystyle \)

\(\displaystyle \)

\(\displaystyle =\int -\sec \theta \ \mathrm d \theta\)

\(=\)

\(\displaystyle -\ln \left \vert {\sec \theta + \tan \theta} \right \vert + C\)

\(\displaystyle \)

\(\displaystyle \)

\(\displaystyle \)

Integral of Secant Function

We have that $\dfrac x a < -1$ for $\theta \in (\pi/2..\pi)$.

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

Which puts $x + \sqrt{x^2 - a^2}$ in the domain of $\ln$ such that $\ln \left \vert \frac x a + \sqrt{\frac {x^2}{a^2} -1} \right \vert < 0$.

So:

\(\displaystyle \)

\(\displaystyle \)

\(\displaystyle \)

\(\displaystyle \)

\(=\)

\(\displaystyle \left \vert {\ln \left \vert{\frac x a + \sqrt {\frac {x^2}{a^2} - 1} } \right \vert} \right \vert + C\)

\(\displaystyle \)

\(\displaystyle \)

\(\displaystyle \)

Definition of Absolute Value

$\blacksquare$