Find the area under the graph of $y=\sqrt{x^{2}-1}$ (the upper half of the right branch of the hyperbola $y^{2}=x^{2}-1$) from 1 to 3. See Figure 9.

Figure 9 $y=\sqrt{x^{2}-1}$, $1 \le x \le 3$

Solution The area $A$ we seek is $A= \int_{1}^{3}\sqrt{x^{2}-1 }\,dx$. The integral contains a square root of the form $\sqrt{x^{2}-a^{2}}$, where $a=1$, so we use the trigonometric substitution $x=\sec \theta$, $0 \le \theta \lt\dfrac{\pi }{2},$ $\pi \le \theta \lt\dfrac{3\pi }{2}.$ Then $dx=\sec \theta \tan \theta \,d\theta$. Since the upper limit $x=3$ does not result in a nice angle $(\theta =\sec ^{-1}\,3)$, we find the indefinite integral first and then use the Fundamental Theorem of Calculus.

With $x=\sec \theta \hbox{ and } dx=\sec \theta \tan \theta d\theta ,$ we have $$\begin{eqnarray*} A&=&\int \sqrt{x^{2}-1}\,dx=\int \sqrt{\sec ^{2}\theta -1}\sec \theta \tan \theta \,d\theta =\int \tan \theta \cdot \sec \theta \tan \theta \,d\theta\\[5pt] &=&\int \tan ^{2}\theta \sec \theta d\theta \end{eqnarray*}$$

Since $\tan \theta$ is raised to an even power and $\sec \theta$ to an odd power, we use the identity $\tan ^{2}\theta =\sec ^{2}\theta -1.$Then $$\begin{eqnarray*} A &=&\int \sqrt{x^{2}-1}\,dx=\int \tan ^{2}\theta \sec \theta d\theta =\int \left( \sec ^{2}\theta -1\right) \sec \theta d\theta\\[4pt] &=&\int \sec ^{3}\theta d\theta -\int \sec \theta d\theta \\[4pt] &=&\dfrac{1}{2}\left[ \sec \theta \tan \theta +\ln \left\vert \sec \theta +\tan \theta \right\vert \right] -\ln \left\vert \sec \theta +\tan \theta \right\vert +C\\[4pt] &=&\dfrac{1}{2}\sec \theta \tan \theta -\dfrac{1}{2}\ln \left\vert \sec \theta +\tan \theta \right\vert +C \end{eqnarray*}$$

Now we express $\tan \theta$ in terms of $x=\sec \theta ,$ and apply the Second Fundamental Theorem of Calculus. $$\begin{eqnarray*} A&=&\int_{1}^{3}\sqrt{x^{2}-1}\, dx\underset{\underset{\underset{\color{#0066A7}{\hbox{\(\tan \theta =\sqrt{x^{2} -1}\)}}}{\color{#0066A7}{\hbox{\(\sec \theta =x\)}}}}{\color{#0066A7}{\uparrow }}} {=}\left[ \frac{1}{2}x\sqrt{x^{2}-1}-\frac{1}{2}\ln \left\vert x+ \sqrt{x^{2}-1}\right\vert \right] _{1}^{3}\\ &=&\dfrac{3}{2}\sqrt{8}-\dfrac{1}{2}\ln ( 3+\sqrt{8}) =3\sqrt{2} -\dfrac{1}{2}\ln (3+2\sqrt{2}) \end{eqnarray*}$$