$I = \int\limits_0^{\frac{\pi }{2}} {\frac{{2 - sinx}}{{2 + cosx}}} dx = \int\limits_0^{\frac{\pi }{2}} {\frac{2}{{2 + cosx}}} dx - \int\limits_0^{\frac{\pi }{2}} {\frac{{sinx}}{{2 + cosx}}} dx$* Tính $J = \int\limits_0^{\frac{\pi }{2}} {\frac{2}{{2 + cosx}}} dx$
Ta có:
\[2 + \cos x = 2 + 2{\cos ^2}\frac{x}{2} - 1 = 2{\cos ^2}\frac{x}{2} + 1 = {\cos ^2}\frac{x}{2}(2 + \frac{1}{{{{\cos }^2}\frac{x}{2}}}) = {\cos ^2}\frac{x}{2}({\tan ^2}\frac{x}{2} + 3)\]
$ \Rightarrow J = \int\limits_0^{\frac{\pi }{2}} {\frac{2}{{{{\cos }^2}\frac{x}{2}({{\tan }^2}\frac{x}{2} + 3)}}} dx$
Đặt $t = \tan \frac{x}{2} \Rightarrow 2dt = \frac{{dx}}{{{{\cos }^2}\frac{x}{2}}}$
$\Rightarrow J = \int\limits_0^1 {\frac{{4dt}}{{{t^2} + 3}}} $
Đặt $t = \sqrt 3 \tan u \Rightarrow dt = \sqrt 3 ({\tan ^2}u + 1)du$
$\Rightarrow J = \int\limits_0^{\frac{\pi }{6}} {\frac{{4\sqrt 3 ({{\tan }^2}u + 1)du}}{{3{{\tan }^2}u + 3}}} = \frac{{4\sqrt 3 }}{3}\int\limits_0^{\frac{\pi }{6}} {du} = \frac{{2\pi \sqrt 3 }}{9}$
* Tính $K = \int\limits_0^{\frac{\pi }{2}} {\frac{{sinx}}{{2 + cosx}}} dx$
Đặt $v = 2 + \cos x \Rightarrow - dv = \sin {\rm{x}}dx$
$\Rightarrow K = \int\limits_2^3 {\frac{{dv}}{v}} = \ln 3 - \ln 2 = \ln \frac{3}{2}$
Vậy \[I = \frac{{2\pi \sqrt 3 }}{9} - \ln \frac{3}{2}\]