Biết \(\int\limits_0^{\frac{\pi }{2}} {\frac{{3\sin x - \cos x}}{{2\sin x + 3\cos x}}dx = \frac{{ - 11}}{{13}}\ln 2 + b\ln 3 + c\pi ,\left( {b,c \in \mathbb{Q}} \right)} \). Giá trị của \(\frac{b}{c}\) là

Hướng dẫn giải

Phân tích \(\frac{{3\sin x - \cos x}}{{2\sin x + 3\cos x}} = \frac{{m\left( {2\sin x + 3\cos x} \right) + n\left( {2\cos x - 3\sin x} \right)}}{{2\sin x + 3\cos x}}\)

                                    \( = \frac{{\left( {2m - 3n} \right)\sin x + \left( {3m + 2n} \right)\cos x}}{{2\sin x + 3\cos x}}\)

Đồng nhất hệ số ta có \(\left\{ \begin{array}{l}2m - 3n = 3\\3m + 2n = - 1\end{array} \right. \Leftrightarrow m = \frac{3}{{13}};n = - \frac{{11}}{{13}}\).

Suy ra \(\int\limits_0^{\frac{\pi }{2}} {\frac{{3\sin x - \cos x}}{{2\sin x + 3\cos x}}dx} = \int\limits_0^{\frac{\pi }{2}} {\frac{{\frac{3}{{13}}\left( {2\sin x + 3\cos x} \right) - \frac{{11}}{{13}}\left( {2\cos x - 3\sin x} \right)}}{{2\sin x + 3\cos x}}} dx.\)

\( = \int\limits_0^{\frac{\pi }{2}} {\left[ {\frac{3}{{13}} - \frac{{11}}{{13}}.\frac{{2\cos x - 3\sin x}}{{2\sin x + 3\cos x}}} \right]dx} = \frac{3}{{13}}\left( x \right)\left| {_{\scriptstyle\atop\scriptstyle0}^{\frac{\pi }{2}}} \right. - \frac{{11}}{{13}}\int\limits_0^{\frac{\pi }{2}} {\frac{{2\cos x - 3\sin x}}{{2\sin x + 3\cos x}}dx.} \)

\( = \frac{{3\pi }}{{26}} - \frac{{11}}{{13}}\int\limits_0^{\frac{\pi }{2}} {\frac{{d\left( {2\sin x + 3\cos x} \right)}}{{2\sin x + 3\cos x}}dx} = \frac{{3\pi }}{{26}} - \frac{{11}}{{13}}\ln \left| {2\sin x + 3\cos x} \right|\left| {_{\scriptstyle\atop\scriptstyle0}^{\frac{\pi }{2}}} \right.\)

\( = \frac{{3\pi }}{{26}} - \frac{{11}}{{13}}\ln 2 + \frac{{11}}{{13}}\ln 3.\) Do đó \(\left\{ \begin{array}{l}b = \frac{{11}}{{13}}\\c = \frac{3}{{26}}\end{array} \right. \Rightarrow \frac{b}{c} = \frac{{11}}{{13}}.\frac{{26}}{3} = \frac{{22}}{3}\)

Chọn A.