Solve $\int_{0}^{1} \min \left\{1, \sqrt{x^{-2}-1}\right\} d x=\log (1+\sqrt{2})$ (cf. Probability and Random Processes by Grimmett) [duplicate]
Solution 1:
You could use $x=\sin\theta, \; dx=\cos\theta d\theta$ to get
$\displaystyle\int_{1/\sqrt{2}}^1 \sqrt{\frac{1}{x^2}-1}dx=\int_{1/\sqrt{2}}^1\frac{\sqrt{1-x^2}}{x}\;dx=\int_{\pi/4}^{\pi/2}\frac{\cos^2\theta}{\sin\theta}d\theta=\int_{\pi/4}^{\pi/2}(\csc\theta-\sin\theta)d\theta$
$\displaystyle=\big[-\ln\big|\csc\theta+\cot\theta\big|+\cos\theta\big]_{\pi/4}^{\pi/2}=\ln(\sqrt{2}+1)-\frac{1}{\sqrt{2}}$.