Question

If ${\int }_1^x\frac{dt}{|t|\sqrt{t^2-1}}=\frac{\pi }{6}$, then $x$ can be equal to :

• A. $\frac{2}{\sqrt{3}}$
• B. $\sqrt{3}$
• C. $2$
• D. $\frac{4}{\sqrt{3}}$

Answer 1

The correct option is A $\frac{2}{\sqrt{3}}$

${\int }_1^x\frac{dt}{\left|t\right|\sqrt{{\left|t\right|}^2-1}}=\frac{\pi }{6}put\left|t\right|=\sec \theta \Rightarrow dt\sec \theta \tan \theta d\theta changinglimit,t=x\Rightarrow \sec \theta =x\Rightarrow {\sec }^{-1}xt=1\Rightarrow \sec \theta =1\Rightarrow {\sec }^{-1}1{\int }_{{\sec }^{-1}1}^{{\sec }^{-1}x}\frac{\sec \theta \tan \theta d\theta }{\sec \theta \sqrt{{\sec }^2\theta -1}}=\frac{\pi }{6}(\because {\sec }^2\theta -1={\tan }^2\theta )\Rightarrow {\int }_{{\sec }^{-1}1}^{{\sec }^{-1}x}\frac{\sec \theta \tan \theta d\theta }{\sec \theta \tan \theta }=\frac{\pi }{6}\Rightarrow {\int }_{{\sec }^{-1}1}^{{\sec }^{-1}x}d\theta =\frac{\pi }{6}{\Rightarrow }_{{\sec }^{-1}1}^{{\sec }^{-1}x}\left[\theta \right]=\frac{\pi }{6}\therefore {\sec }^{-1}x-{\sec }^{-1}1=\frac{\pi }{6}\therefore {\sec }^{-1}x=\frac{\pi }{6}+{\sec }^{-1}1=\frac{\pi }{6}+0\therefore x=\sec \frac{\pi }{6}=\frac{2}{\sqrt{3}}$