Cho tích phân \(I = \int\limits_0^\pi {\frac{{x\sin x}}{{{{\cos }^2}x – 16}}} dx = a\pi \ln \frac{b}{c},\,\left( {a,b,c \in \mathbb{Q},0 < b < c < 8} \right)\). Giá trị của biểu thức \(40a + 3b – {c^2}\)là
A.\(17\).
B.\(13\).
C.\( – 9\).
D.\( – 11\).
Lời giải:
Đặt \(t = \pi – x \Leftrightarrow x = \pi – t \Rightarrow dx = – dt\). Đổi cận \(\left\{ \begin{array}{l}x = 0 \Rightarrow t = \pi \\x = \pi \Rightarrow t = 0\end{array} \right.\)
\(\begin{array}{l}I = – \int\limits_\pi ^0 {\frac{{\left( {\pi – t} \right)\sin \left( {\pi – t} \right)}}{{{{\cos }^2}\left( {\pi – t} \right) – 16}}dt = \int\limits_0^\pi {\frac{{\left( {\pi – t} \right)\sin t}}{{{{\cos }^2}t – 16}}dt = \pi \int\limits_0^\pi {\frac{{\sin t}}{{{{\cos }^2}t – 16}}dt – } } } \int\limits_0^\pi {\frac{{t\sin t}}{{{{\cos }^2}t – 16}}} dt\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \pi \int\limits_0^\pi {\frac{{\sin x}}{{{{\cos }^2}x – 16}}dx – } \int\limits_0^\pi {\frac{{x\sin x}}{{{{\cos }^2}x – 16}}} dx = \pi \int\limits_0^\pi {\frac{{\sin x}}{{{{\cos }^2}x – 16}}dx – } I\end{array}\)
Suy ra \(2I = \pi \int\limits_0^\pi {\frac{{\sin x}}{{{{\cos }^2}x – 16}}dx} \)
\( \Leftrightarrow I = – \frac{\pi }{2}\int\limits_0^\pi {\frac{{d\left( {\cos x} \right)}}{{{{\cos }^2}x – 16}}} = \frac{\pi }{{16}}\left[ {\int\limits_0^\pi {\frac{{d\left( {\cos x} \right)}}{{\cos x + 4}} – \int\limits_0^\pi {\frac{{d\left( {\cos x} \right)}}{{\cos x – 4}}} } } \right] = \frac{\pi }{{16}}\ln \left| {\frac{{\cos x + 4}}{{\cos x – 4}}} \right|\left| \begin{array}{l}\pi \\0\end{array} \right. = \frac{1}{8}\pi \ln \frac{3}{5}\)
Do đó \(a = \frac{1}{8};b = 3,c = 5 \Rightarrow 40a + 3b – {c^2} = – 11\).
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