Biết tích phần từ 0 đến pi/2 3 sin x + cos x / 2 sin x + 3 cos x d x = − 7/13 ln 2 + b ln 3 + cpi ( b , c thuộc Q ) . . Tính b/c .

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

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

\( \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{2A - 3B = 3}\\{3A + 2B = 1}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{A = \frac{9}{{13}}}\\{B = - \frac{7}{{13}}}\end{array}} \right.\)

Nên

\[3\sin x + \cos x = \frac{9}{{13}}\left( {2\sin x + 3\cos x} \right) - \frac{7}{{13}}\left( {2\cos x - 3\sin x} \right)\]

Từ đó ta có

\(\int\limits_0^{\frac{\pi }{2}} {\frac{{3sinx + cosx}}{{2sinx + 3cosx}}} dx = \int\limits_0^{\frac{\pi }{2}} {\frac{{\frac{9}{{13}}(2sinx + 3cosx) - \frac{7}{{13}}(2cosx - 3sinx)}}{{2sinx + 3cosx}}dx} \)

\(\begin{array}{l} = \frac{9}{{13}}\int\limits_0^{\frac{\pi }{2}} {dx - \frac{7}{{13}}} \int\limits_0^{\frac{\pi }{2}} {\frac{{2cosx - 3sinx}}{{2sinx + 3cosx}}} dx\\ = \frac{{9\pi }}{{26}} - \frac{7}{{13}}\int\limits_0^{\frac{\pi }{2}} {\frac{1}{{2sinx + 3cosx}}} d(2sinx + 3cosx)\\ = \frac{{9\pi }}{{26}} - \frac{7}{{13}}ln|2sinx + 3cosx|\left| {_0^{\frac{\pi }{2}}} \right.\\ = \frac{{9\pi }}{{26}} - \frac{7}{{13}}ln2 + \frac{7}{{13}}ln3\end{array}\)

Suy ra\[b = \frac{7}{{13}};c = \frac{9}{{26}} \Rightarrow \frac{b}{c} = \frac{{14}}{9}\]