Exercise:
Compute, $\int_{\sqrt{2}}^{2}{\frac{dx}{{{x}^{2}}\sqrt{{{x}^{2}}-1}}}$
Solution:
Let$x=\sec u\Rightarrow dx=\sec u\tan u\,du$
So $\int_{\sqrt{2}}^{2}{\frac{dx}{{{x}^{2}}\sqrt{{{x}^{2}}-1}}=\int_{x=\sqrt{2}}^{x=2}{\frac{\sec
u\tan udu}{{{\sec }^{2}}u\sqrt{{{\sec }^{2}}u-1}}}}$
$=\int_{x=\sqrt{2}}^{x=2}{\frac{\sec u\tan
udu}{{{\sec }^{2}}u\tan u}}$ $=\int_{x=\sqrt{2}}^{x=2}{\frac{du}{\sec u}}$ $=\int_{x=\sqrt{2}}^{x=2}{\cos
u\,du}$ $=\left[ \sin u \right]_{x=\sqrt{2}}^{x=2}$
but $x=\frac{1}{\cos
u}\Leftrightarrow \sin u=\sqrt{1-{{\cos }^{2}}u}=\sqrt{1-\frac{1}{{{x}^{2}}}}$
$\int_{\sqrt{2}}^{2}{\frac{dx}{{{x}^{2}}\sqrt{{{x}^{2}}-1}}}$$=\left[
\sqrt{1-\frac{1}{{{x}^{2}}}}
\right]_{x=\sqrt{2}}^{x=2}=\sqrt{1-\frac{1}{4}}-\sqrt{1-\frac{1}{2}}$ $=\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}=\frac{\sqrt{3}-\sqrt{2}}{2}$