$\int\limits_0^{\pi /2} {\cfrac{{{{\cos }^2}xdx}}{{{{\cos }^2}x + 4{{\sin }^2}x}}}$

Let $I = \int\limits_0^{\pi /2} {\cfrac{{{{\cos }^2}x}}{{{{\cos }^2}x + 4{{\sin }^2}x}}} dx$

$= \int\limits_0^{\pi /2} {\cfrac{{{{\cos }^2}x}}{{{{\cos }^2}x + 4\left( {1 - {{\cos }^2}x} \right)}}dx}$

$= \int\limits_0^{\pi /2} {\cfrac{{{{\cos }^2}x}}{{4 - 3{{\cos }^2}x}}dx} = - \cfrac{1}{3}\int\limits_0^{\pi /2} {\cfrac{{4 - 3{{\cos }^2}x - 4}}{{4 - 3{{\cos }^2}x}}} dx$

$= - \cfrac{1}{3}\int\limits_0^{\pi /2} {\left( {1 - \cfrac{4}{{4 - 3{{\cos }^2}x}}} \right)dx = - \cfrac{1}{3}\int\limits_0^{\pi /2} {dx + \cfrac{4}{3}\int\limits_0^{\pi /2} {\cfrac{{dx}}{{4 - 3{{\cos }^2}x}}} } }$

$= - \cfrac{1}{3}\left( {\cfrac{\pi }{2}} \right) + \cfrac{4}{3}\int\limits_0^{\pi /2} {\cfrac{{{{\sec }^2}x}}{{4{{\sec }^2}x - 3}}dx = - \cfrac{\pi }{6} + \cfrac{4}{3}\int\limits_0^{\pi /2} {\cfrac{{{{\sec }^2}x}}{{4\left( {1 + {{\tan }^2}x} \right) - 3}}} dx}$

Put $\tan x = t$ $\Rightarrow$ ${\sec ^2}xdx = dt$
When $x = 0,t = 0$ and when $x = \cfrac{\pi }{2},t = \infty$

$\therefore$ $I = - \cfrac{\pi }{6} + \cfrac{4}{3}\int\limits_0^\infty {\cfrac{{dt}}{{4\left( {1 + {t^2}} \right) - 3}}} = \cfrac{\pi }{6} + \cfrac{4}{3}\int\limits_0^\infty {\cfrac{{dt}}{{4{t^2} + 1}}}$

$= - \cfrac{\pi }{6} + \cfrac{4}{3} \cdot \cfrac{1}{4}\int\limits_0^\infty {\cfrac{{dt}}{{{t^2} + \cfrac{1}{4}}}} = - \cfrac{\pi }{6} + \cfrac{1}{3} \cdot \cfrac{1}{{1/2}}\left[ {{{\tan }^{ - 1}}\cfrac{t}{{1/2}}} \right]_0^\infty$

$= - \cfrac{\pi }{6} + \cfrac{2}{3}\left[ {{{\tan }^{ - 1}}2t} \right]_0^\infty = - \cfrac{\pi }{6} + \cfrac{2}{3}\left[ {{{\tan }^{ - 1}}\infty - {{\tan }^{ - 1}}0} \right]$

$= - \cfrac{\pi }{6} + \cfrac{2}{3}\left[ {\cfrac{\pi }{2} - 0} \right] = - \cfrac{\pi }{6} + \cfrac{\pi }{3} = \cfrac{\pi }{6}$