Biết \[\mathop \smallint \limits_0^{\frac{\pi }{2}} \frac{{3\sin x + \cos x}}{{2\sin x + 3\cos x}}dx = - \frac{7}\ln 2 + b\ln 3 + c\pi \,\,\left( {b,c \in \mathbb{Q}} \right).\]. Tính \(\frac{b}{c}\).

Ta có\[3\sin x + \cos x = A\left( {2\sin x + 3\cos x} \right) + B\left( {2\cos x - 3\sin x} \right)\]

\[ \Leftrightarrow 3sinx + cosx = (2A - 3B)sinx + (3A + 2B)cosx\]

\( \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{2A - 3B = 3}\\{3A + 2B = 1}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{A = \frac{9}}\\{B = - \frac{7}}\end{array}} \right.\)

Nên

\[3\sin x + \cos x = \frac{9}\left( {2\sin x + 3\cos x} \right) - \frac{7}\left( {2\cos x - 3\sin x} \right)\]

Từ đó ta có

\(\int\limits_0^{\frac{\pi }{2}} {\frac} dx = \int\limits_0^{\frac{\pi }{2}} {\frac{{\frac{9}(2sinx + 3cosx) - \frac{7}(2cosx - 3sinx)}}dx} \)

\(\begin{array}{l} = \frac{9}\int\limits_0^{\frac{\pi }{2}} {dx - \frac{7}} \int\limits_0^{\frac{\pi }{2}} {\frac} dx\\ = \frac{{9\pi }} - \frac{7}\int\limits_0^{\frac{\pi }{2}} {\frac{1}} d(2sinx + 3cosx)\\ = \frac{{9\pi }} - \frac{7}ln|2sinx + 3cosx|\left| {_0^{\frac{\pi }{2}}} \right.\\ = \frac{{9\pi }} - \frac{7}ln2 + \frac{7}ln3\end{array}\)

Suy ra\[b = \frac{7};c = \frac{9} \Rightarrow \frac{b}{c} = \frac{9}\]

Đáp án cần chọn là: B