Question

\[\int\limits_0^{\pi/2} \frac{1}{2 + \cos x} dx\] equals

Options

• \[\frac{1}{3} \tan^{- 1} \left( \frac{1}{\sqrt{3}} \right)\]
• \[\frac{2}{\sqrt{3}} \tan^{- 1} \left( \frac{1}{\sqrt{3}} \right)\]
• \[\sqrt{3} \tan^{- 1} \left( \sqrt{3} \right)\]

   

• \[2\sqrt{3} \tan^{- 1} \sqrt{3}\]

Answer 1

\[\ \frac{2}{\sqrt{3}} \tan^{- 1} \left( \frac{1}{\sqrt{3}} \right)\]

\[\text{We have}, \]

\[I = \int_0^\frac{\pi}{2} \frac{1}{2 + \cos x} d x\]

\[ = \int_0^\frac{\pi}{2} \frac{1}{2 + \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}}} d x\]

\[ = \int_0^\frac{\pi}{2} \frac{1 + \tan^2 \frac{x}{2}}{2 + 2 \tan^2 \frac{x}{2} + 1 - \tan^2 \frac{x}{2}}dx\]

\[ = \int_0^\frac{\pi}{2} \frac{\sec^2 \frac{x}{2}}{3 + \tan^2 \frac{x}{2}}dx\]

\[\text{Putting} \tan \frac{x}{2} = t\]

\[ \Rightarrow \frac{1}{2} \sec^2 \frac{x}{2}dx = dt\]

\[ \Rightarrow \sec^2 \frac{x}{2}dx = 2dt\]

\[When, x \to 0; t \to 0\]

\[and x \to \frac{\pi}{2}; t \to 1\]

\[ \therefore I = \int_0^1 \frac{2}{3 + t^2}dt\]

\[ = 2 \int_0^1 \frac{1}{\left( \sqrt{3} \right)^2 + t^2}dt\]

\[ = \frac{2}{\sqrt{3}} \left[ \tan^{- 1} \frac{t}{\sqrt{3}} \right]_0^1 \]

\[ = \frac{2}{\sqrt{3}}\left[ \tan^{- 1} \frac{1}{\sqrt{3}} - \tan^{- 1} \frac{0}{\sqrt{3}} \right]\]

\[ = \frac{2}{\sqrt{3}} \tan^{- 1} \left( \frac{1}{\sqrt{3}} \right)\]