Cho \(\int\limits_0^{\frac{\pi }{4}} {\frac{{\ln \left( {\sin x + 2\cos x} \right)}}{{{{\cos }^2}x}}} dx = a\ln 3 + b\ln 2 + c\pi \) với \(a,b,c\) là các số hữu tỉ.

Giá trị của abc bằng

Hướng dẫn giải

Đặt \(\left\{ \begin{array}{l}u = \ln \left( {\sin x + 2\cos x} \right)\\dv = \frac{{dx}}{{{{\cos }^2}x}}\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = \frac{{\cos x - 2\sin x}}{{\sin x + 2\cos x}}dx\\v = \tan x + 2\end{array} \right..\)

Khi đó

\(\int\limits_0^{\frac{\pi }{4}} {\frac{{\ln \left( {\sin x + 2\cos x} \right)}}{{{{\cos }^2}x}}} dx = \left( {\tan x + 2} \right)\ln \left( {\sin x + 2\cos x} \right)\left| {_{\scriptstyle\atop\scriptstyle0}^{\frac{\pi }{4}}} \right. - \int\limits_0^{\frac{\pi }{4}} {\frac{{\cos x - 2\sin x}}{{\cos x}}dx} \)

                                    \( = 3\ln \left( {\frac{{3\sqrt 2 }}{2}} \right) - 2\ln 2 - \int\limits_0^{\frac{\pi }{4}} {\left( {1 - 2\tan x} \right)} dx\)

                                    \( = 3\ln 3 - \frac{7}{2}\ln 2 - \left( {x + 2\ln \left| {\cos x} \right|} \right)\left| {_{\scriptstyle\atop\scriptstyle0}^{\frac{\pi }{4}}} \right.\)

                                    \( = 3\ln 3 - \frac{7}{2}\ln 2 - \frac{\pi }{4} - 2\ln \frac{{\sqrt 2 }}{2} = 3\ln 3 - \frac{5}{2}\ln 2 - \frac{\pi }{4}.\)

Suy ra \(a = 3,b = - \frac{5}{2},c = - \frac{1}{4}.\) Vậy \(abc = 18.\)

Chọn A.