Question

$\underset{2}{\overset{4}{\int }}\frac{\sqrt{x^2-4}}{x}dx$ equal to

• A. $2(3\sqrt{3}-\pi )$
• B. $2\sqrt{3}-\pi$
• C. $\frac{2}{3}(3\sqrt{3}-\pi )$
• D. $\pi$

Answer 1

Answer: (C)

Put $x=2\sec \theta$
$\Rightarrow \underset{2}{\overset{4}{\int }}\frac{\sqrt{x^2-4}}{x}dx=\underset{0}{\overset{\pi /3}{\int }}\frac{2\tan \theta }{2\sec \theta }\cdot 2\sec \theta \tan \theta d\theta$
$=\underset{0}{\overset{\pi /3}{\int }}2{\tan }^2\theta d\theta =2\underset{0}{\overset{\pi /3}{\int }}\left({\sec }^2\theta -1\right)d\theta$
$=2(\tan \theta -\theta ){\Vert }_0^{\pi /3}=2\left(\left(\sqrt{3}-\frac{\pi }{3}\right)-(0-0)\right)=\frac{2}{3}(3\sqrt{3}-\pi )$