Finding an Integral Containing \(\sqrt{x^{2}-4}\)

Find \(\int \frac{\sqrt{x^{2}-4}}{x} dx\).

Solution The integrand contains the square root \(\sqrt{x^{2}-4}\) that is of the form \(\sqrt{x^{2}-a^{2}},\) where \(a=2\). We use the substitution \(x=2\sec \theta\), \(0 \le \theta \lt \dfrac{\pi }{2},\) \(\pi \le \theta \lt\dfrac{3\pi }{2}\). Then \(dx=2\sec \theta \tan \theta~d\theta.\) Since \[ \begin{eqnarray*} \sqrt{x^{2}-4} \underset{\underset{\color{#0066A7}{\hbox{\(x=2 \sec \theta\)}}}{\color{#0066A7}{\uparrow}}}{=}\sqrt{4\sec ^{2}\theta -4}=2\sqrt{\sec ^{2}\theta -1}=2\sqrt{\tan^{2}\theta } \underset{\underset{\underset{\color{#0066A7}{\hbox{since \(0 \le \theta \lt \dfrac{\pi }{2}, \pi \le \theta \lt \dfrac{3\pi}{2}\)}}}{\color{#0066A7}{\hbox{\(\tan \theta \ge 0\)}}}}{\color{#0066A7}{\uparrow}}}{=} 2\tan \theta \end{eqnarray*} \]

we have \[ \begin{eqnarray*} \int \frac{\sqrt{x^{2}-4}}{x}dx&=&\int \frac{(2\tan \theta )(2\sec \theta \tan \theta \,d\theta )}{2\sec \theta }=2\int \tan ^{2}\theta d\theta \underset{\underset{\color{#0066A7}{\hbox{\(\tan^{2}\theta = \sec^{2}\theta -1\)}}}{\color{#0066A7}{\uparrow}}}{=}2\int (\sec^{2}\theta -1) d\theta \\ &=&2\int \sec^{2}\theta d\theta -2\int d\theta =2\tan \theta -2\theta +C \end{eqnarray*} \]

To express the solution in terms of \(x\), we use either the right triangles in Figure 7 or trigonometric identities.

Using identities, we find, \[ \begin{eqnarray*} \tan \theta \underset{\underset{\underset{{\color{#0066A7}{\hbox{\(\tan \theta \ge 0\)}}}}{\color{#0066A7}{\hbox{\(\tan^{2}\theta =\sec^{2}\theta -1\)}}}}{\color{#0066A7}{\uparrow}}}{=}\sqrt{\sec^{2}\theta -1}= \sqrt{\frac{x^{2}}{4}-1}=\frac{1}{2}\sqrt{x^{2}-4} \end{eqnarray*} \]

Also since \(\sec \theta =\dfrac{x}{2},\) \(0 \le \theta \lt \dfrac{\pi }{2},\) \(\pi \le \theta \lt \dfrac{3\pi }{2},\) the inverse function \(\theta =\sec ^{-1}\dfrac{x }{2}\) is defined.

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Then \[ \int\frac{\sqrt{x^{2}-4}}{x}dx=2\tan \theta -2\theta+C=\sqrt{x^{2}-4} -2\sec ^{-1} \frac{x}{2}+C \]