Question

If \(\displaystyle \int_{1}^{x}\frac{dt}{t\sqrt{t^2-1}}=\frac{\pi}{6}\), then \(x\) can be equal to

• A. \(\dfrac{2}{\sqrt{3}}\)
• B. \(\sqrt{3}\)
• C. 2
• D. None of these

Hint:

\(\int \frac{dx}{x\sqrt{x^2-1}}=\sec^{-1}(x)+C\). In definite form, apply directly and solve for \(x\) using inverse secant.

Answer 1

The Correct Option is A

Step 1: Recognize standard integral form.
\[ \int \frac{dt}{t\sqrt{t^2-1}} = \sec^{-1}(t)+C \] (for \(t\ge 1\)).
Step 2: Apply limits.
\[ \int_{1}^{x}\frac{dt}{t\sqrt{t^2-1}} = \sec^{-1}(x)-\sec^{-1}(1) \] Step 3: Evaluate \(\sec^{-1}(1)\).
\[ \sec^{-1}(1)=0 \] So:
\[ \sec^{-1}(x)=\frac{\pi}{6} \] Step 4: Convert to \(\sec\).
\[ x=\sec\left(\frac{\pi}{6}\right) =\frac{1}{\cos\left(\frac{\pi}{6}\right)} =\frac{1}{\frac{\sqrt{3}}{2}} =\frac{2}{\sqrt{3}} \] Final Answer: \[ \boxed{\frac{2}{\sqrt{3}}} \]