If $\displaystyle \int \limits_{\sqrt{2}}^{k} \dfrac{dx}{x \sqrt{x^2 - 1}} \; dx = \dfrac{\pi}{12}$, obtain k.
Let $I = \displaystyle \int \limits_{\sqrt{2}}^{k} \dfrac{dx}{x \sqrt{x^2 - 1}} \; dx$ $\;\;\; \cdots \; (1)$
Put $x = \sec \theta$ $\;\;\; \cdots \; (2)$
Differentiating equation $(2)$ gives
$dx = \sec \theta \tan \theta \; d\theta$ $\;\;\; \cdots \; (2a)$
From equation $(2)$, $\theta = \sec^{-1}x$
$\therefore \; \text{When } \begin{cases}
x = \sqrt{2}, & \theta = \sec^{-1} \left(\sqrt{2}\right) = \dfrac{\pi}{4} \\
x = k, & \theta = \sec^{-1} \left(k\right)
\end{cases}$ $\;\;\; \cdots \; (2b)$
$\therefore$ $\;$ In view of equations $(2)$, $(2a)$ and $(2b)$, equation $(1)$ can be written as
$\begin{aligned}
I & = \int \limits_{\frac{\pi}{4}}^{\sec^{-1} \left(k\right)} \dfrac{\sec \theta \; \tan \theta \; d \theta}{\sec \theta \; \sqrt{\sec^2 \theta - 1}} \\\\
& = \int \limits_{\frac{\pi}{4}}^{\sec^{-1} \left(k\right)} d \theta \\\\
& = \left[\theta\right]_{\pi / 4}^{\sec^{-1} \left(k\right)} \\\\
& = \sec^{-1} \left(k\right) - \dfrac{\pi}{4} \;\;\; \cdots \; (3)
\end{aligned}$
Given $I = \dfrac{\pi}{12}$ $\;\;\; \cdots \; (4)$
$\therefore$ $\;$ We have from equations $(3)$ and $(4)$,
$\sec^{-1} \left(k\right) - \dfrac{\pi}{4} = \dfrac{\pi}{12}$
i.e. $\sec^{-1} \left(k\right) = \dfrac{\pi}{12} + \dfrac{\pi}{4} = \dfrac{\pi}{3}$
$\implies$ $k = \sec \left(\dfrac{\pi}{3}\right) = 2$