<p>Evaluate \(\displaystyle\int_1^\infty\frac{dx}{x\sqrt{x^2-1}}\) [JEE Advanced 2003]</p>
Step-by-Step Solution
Key Concept: Let x = sec \theta: dx = sec \theta \cdot tan \theta d\theta, \sqrt{x^2-1} = tan \theta. Integral \to \int_0^(\pi/2) d\theta = \pi/2.
<div class='solution'>
<p>Let $x=\sec\theta$, $\sqrt{x^2-1}=\tan\theta$, $dx=\sec\theta\tan\theta\,d\theta$. Limits: $x=1\to\theta=0$; $x\to\infty\to\theta\to\pi/2$.</p>
<p>$$\int_1^\infty\frac{dx}{x\sqrt{x^2-1}}=\int_0^{\pi/2}\frac{\sec\theta\tan\theta\,d\theta}{\sec\theta\tan\theta}=\int_0^{\pi/2}d\theta=\boxed{\frac{\pi}{2}}$$</p>
</div>
Correct Answer: A