Definite Integration Question 326
Question: $ \int_0^{\pi /2}{\frac{dx}{2+\cos x}}= $
[BIT Ranchi 1992]
Options:
A) $ \frac{1}{\sqrt{3}}{{\tan }^{-1}}( \frac{1}{\sqrt{3}} ) $
B) $ \sqrt{3}{{\tan }^{-1}}( \sqrt{3} ) $
C) $ \frac{2}{\sqrt{3}}{{\tan }^{-1}}( \frac{1}{\sqrt{3}} ) $
D) $ 2\sqrt{3}{{\tan }^{-1}}( \sqrt{3} ) $
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Answer:
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}}( \frac{1}{\sqrt{3}} )} $ .
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