Let I = \(\int\limits^{\frac{\pi}{2}}_{0} \frac{1}{2 + cos \,x }dx\)
\(= \int\limits^{\frac{\pi}{2}}_{0} \frac{1}{2 + \cfrac{1 - tan^2\frac x2}{1+ tan^2\frac x2}}dx\) \(\left(\because cos \, x = \frac{1- tan^2\frac{x}{2}}{1+tan^2\,\frac x2}\right)\)
\(= \int\limits^{\frac{\pi}{2}}_{0}{ \cfrac{1 +tan^2\frac x2}{tan^2\frac x2 +3}}dx\)
\(= \int\limits^{\frac{\pi}{2}}_{0}{ \cfrac{sec^2\frac x2}{tan^2\frac x2 +3}}dx\)
Let \(tan \frac x2=t\)
Then\(\frac12 sec^2 \frac x2 dx = dt\)
⇒ \(sec^2 \frac x2 dx = 2dt\)
Limit converts from t = tan 0 = 0 to t = \(tan \frac {\pi}4=1\)
\(\therefore I = \int\limits^1_0 \frac{2dt}{t^2 +3}\)
\(= \frac2{\sqrt3} \left(tan^{-1}\frac t {\sqrt 3}\right)^1_0\)
\(= \frac2{\sqrt3} \left(tan^{-1}\left(\frac 1{\sqrt 3}\right) - tan^{-1}0\right)\)
\(= \frac2{\sqrt3} \left(\frac{\pi}{6} - 0\right)\)
\(= \frac{\pi}{3\sqrt3}\)
