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