A) \[\frac{1}{\sqrt{3}}{{\tan }^{-1}}\left( \frac{1}{\sqrt{3}} \right)\]
B) \[\sqrt{3}{{\tan }^{-1}}\left( \sqrt{3} \right)\]
C) \[\frac{2}{\sqrt{3}}{{\tan }^{-1}}\left( \frac{1}{\sqrt{3}} \right)\]
D) \[2\sqrt{3}{{\tan }^{-1}}\left( \sqrt{3} \right)\]
Correct Answer: C
Solution :
\[I=\int_{0}^{\pi /2}{\frac{dx}{2+\cos x}}\] \[=\int_{0}^{\pi /2}{\frac{dx}{2{{\sin }^{2}}\frac{x}{2}+2{{\cos }^{2}}\frac{x}{2}+{{\cos }^{2}}\frac{x}{2}-{{\sin }^{2}}\frac{x}{2}}}\] \[=\int_{0}^{\pi /2}{\frac{dx}{{{\sin }^{2}}\frac{x}{2}+3{{\cos }^{2}}\frac{x}{2}}}=\int_{0}^{\pi /2}{\frac{{{\sec }^{2}}\frac{x}{2}}{3+{{\tan }^{2}}\frac{x}{2}}dx}\] Put \[t=\tan \frac{x}{2}\Rightarrow dt=\frac{1}{2}{{\sec }^{2}}\frac{x}{2}dx\], then \[I=2\int_{0}^{1}{\frac{dt}{3+{{t}^{2}}}=\frac{2}{\sqrt{3}}{{\tan }^{-1}}\left( \frac{1}{\sqrt{3}} \right)}\].
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