If $\int\limits_1^x\frac{dt}{|t|\sqrt{t^2-1}}=\frac{π}{6}$, then x can be equal to:
Answer & explanation
Correct answer: option 1
$\int\limits_1^x\frac{dt}{|t|\sqrt{t^2-1}}=\frac{π}{6}⇒[sec^{-1}(t)]_1^x=\frac{π}{6}⇒sec^{-1}(x)=\frac{π}{6}$
$⇒sec^{-1}(x)=\frac{π}{6}⇒x=\frac{2}{\sqrt{3}}$