A) \[2\sqrt{3}+\frac{2\pi }{3}\]
B) \[2\sqrt{3}-\frac{\pi }{3}\]
C) \[2\sqrt{3}+\frac{\pi }{3}\]
D) \[2\sqrt{3}-\frac{2\pi }{3}\]
Correct Answer: D
Solution :
\[I=\int_{2}^{4}{\frac{\sqrt{{{x}^{2}}-4}}{x}}dx\] Let \[x=2\sec \theta \Rightarrow dx=2\sec \theta \tan \theta d\theta \] When\[x=2,\theta =0\]when \[x=4,\text{ }\theta =\frac{\pi }{3}\] \[\therefore \] \[I=\int_{0}^{\pi /3}{\frac{\sqrt{4{{\sec }^{2}}\theta -4}}{2\sec \theta }}=2\tan \theta \sec \theta d\theta \] \[=\int_{0}^{\pi /3}{2\tan \theta .\tan \theta \,d\theta }\] \[=2\int_{0}^{\pi /3}{{{\tan }^{2}}\theta .d\theta }\] \[=2\int_{0}^{\pi /3}{(se{{c}^{2}}\theta -1)d\theta }\] \[=2[\tan \theta -\theta ]_{0}^{\pi /3}\] \[=2\left[ \tan \frac{\pi }{3}-\frac{\pi }{3}-\tan \theta \right]\] \[=2\left[ \sqrt{3}-\frac{\pi }{3}-0 \right]=2\sqrt{3}-\frac{2\pi }{3}\]
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