Cho \(\int\limits_0^{\frac{\pi }{2}} {\frac{{\cos x}}{{{{\sin }^2}x - 5\sin x + 6}}dx} = a\ln \frac{4}{c} + b\,\,\left( {c > 0} \right)\)

Cho \(\int\limits_0^{\frac{\pi }{2}} {\frac{{\cos x}}{{{{\sin }^2}x - 5\sin x + 6}}dx} = a\ln \frac{4}{c} + b\,\,\left( {c > 0} \right)\) . Tính tổng a + b + c?
A. 3
B. 4
C. 0
D. 1

\(I = \int\limits_0^{\frac{\pi }{2}} {\frac{{\cos x}}{{{{\left( {\sin x} \right)}^2} - 5\sin x + 6}}dx} \)
Đặt \(t = \sin x \Rightarrow dt = \cos xdx\)
Đổi cận: \(\left\{ \begin{array}{l}x = 0 \Rightarrow t = 0\\x = \frac{\pi }{2} \Rightarrow t = 1\end{array} \right.\)
\(I = \int\limits_0^1 {\frac{1}{{{t^2} - 5t + 6}}dt} = \int\limits_0^1 {\frac{1}{{(t - 2)(t - 3)}}dt} = \int\limits_0^1 {\frac{1}{{(t - 2)(t - 3)}}dt} = \int\limits_0^1 {\left( {\frac{1}{{t - 3}} - \frac{1}{{t - 2}}} \right)dt} \)
\( = \ln \left| {\frac{{t - 3}}{{t - 2}}} \right|_0^1 = \ln 2 - \ln \frac{3}{2} = \ln \frac{4}{3}.\)
Do đó: a = 1; b = 0; c = 3
S = a + b + c = 1 + 0 + 3 = 4.