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If $\int_{1}^{x} \frac{d t}{|t| \sqrt{t^{2}-1}}=\frac{\pi}{6}$, then $\mathrm{x}$ can be equal to
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Verified Answer
The correct answer is:
$\frac{2}{\sqrt{3}}$
$$\begin{array}{l} \int_{1}^{x} \frac{d t}{|t| \sqrt{t^{2}-1}}=\frac{\pi}{6} \\
\Rightarrow\left[\mathrm{sec}^{-1} t\right]_{1}^{x}=\frac{\pi}{6} \\
\Rightarrow \sec ^{-1} x-\sec ^{-1} 1=\frac{\pi}{6} \\
\Rightarrow \sec ^{-1} x-0=\frac{\pi}{6} \Rightarrow x=\sec \frac{\pi}{6} \\
\Rightarrow x=\frac{2}{\sqrt{3}}
\end{array}
$$
\Rightarrow\left[\mathrm{sec}^{-1} t\right]_{1}^{x}=\frac{\pi}{6} \\
\Rightarrow \sec ^{-1} x-\sec ^{-1} 1=\frac{\pi}{6} \\
\Rightarrow \sec ^{-1} x-0=\frac{\pi}{6} \Rightarrow x=\sec \frac{\pi}{6} \\
\Rightarrow x=\frac{2}{\sqrt{3}}
\end{array}
$$
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