Answer
$$\int^2_{2/\sqrt3}\frac{\cos(\sec^{-1}x)dx}{x\sqrt{x^2-1}}=\frac{\sqrt3-1}{2}$$
Work Step by Step
$$A=\int^2_{2/\sqrt3}\frac{\cos(\sec^{-1}x)dx}{x\sqrt{x^2-1}}$$
We set $u=\sec^{-1}x$, which means $$du=\frac{dx}{x\sqrt{x^2-1}}$$
- For $x=\frac{2}{\sqrt3}$, we have $u=\sec^{-1}\Big(\frac{2}{\sqrt3}\Big)=\cos^{-1}\Big(\frac{\sqrt3}{2}\Big)=\frac{\pi}{6}$
- For $x=2$, we have $u=\sec^{-1}2=\cos^{-1}\Big(\frac{1}{2}\Big)=\frac{\pi}{3}$
Therefore, $$A=\int^{\pi/3}_{\pi/6}\cos udu$$ $$A=\sin u\Big]^{\pi/3}_{\pi/6}$$ $$A=\sin\frac{\pi}{3}-\sin\frac{\pi}{6}$$ $$A=\frac{\sqrt3}{2}-\frac{1}{2}=\frac{\sqrt3-1}{2}$$