2/ $I = \int\limits_0^{\pi /2} {\frac{{{{\cos }^2}x}}{{3\sin
x + 4\cos x}}} dx$
Đặt $J = \int\limits_0^{\pi /2} {\frac{{{{\sin
}^2}x}}{{3\sin x + 4\cos x}}} dx$
* $16I - 9J = \int\limits_0^{\pi /2} {\frac{{16{{\cos }^2}x
- 9{{\sin }^2}x}}{{3\sin x + 4\cos x}}} dx = \int\limits_0^{\pi /2} {(4\cos x -
3\sin x)} dx = 1 (1)$.
* $I + J = \int\limits_0^{\pi /2} {\frac{1}{{3\sin x + 4\cos
x}}} dx$
$3\sin x + 4\cos x = 6\sin \frac{x}{2}\cos \frac{x}{2} +
4{\cos ^2}\frac{x}{2} - 4{\sin ^2}\frac{x}{2}$
$ = - 2{\cos
^2}\frac{x}{2}\left( {2{{\tan }^2}\frac{x}{2} - 3\tan \frac{x}{2} - 2} \right)$
Đặt $t = \tan \frac{x}{2} \Rightarrow dt =
\frac{{dx}}{{2{{\cos }^2}\frac{x}{2}}}$
$I + J = -
\int\limits_0^{\pi /2} {\frac{{dx}}{{2{{\cos }^2}\frac{x}{2}\left( {2{{\tan
}^2}\frac{x}{2} - 3\tan \frac{x}{2} - 2} \right)}}} = -
\int\limits_0^1 {\frac{{dt}}{{2{t^2} - 3t - 2}}}$
$ = - \frac{1}{5}\int\limits_0^1 {\left(
{\frac{1}{{t - 2}} - \frac{2}{{2t + 1}}} \right)dt = \frac{1}{5}} \ln 6 (2).$
* Từ (1) và (2) suy ra $I = \frac{{5 + 9\ln 6}}{{125}}$