Chọn phương án B.

Đặt \(t=\sin x\) ta có

• \(\mathrm{d}t=\cos x\mathrm{\,d}x\)
• \(x=0\Rightarrow t=0\)
• \(x=\dfrac{\pi}{2}\Rightarrow t=1\)

Khi đó $$\begin{aligned}
\displaystyle\int\limits_{0}^{\tfrac{\pi}{2}}\dfrac{\cos x}{\left(\sin x\right)^2-5\sin x+6}\mathrm{\,d}x&=\displaystyle\int\limits_{0}^{1}\dfrac{1}{t^2-5t+6}\mathrm{\,d}t\\
&=\displaystyle\int\limits_{0}^{1}\dfrac{1}{(t-2)(t-3)}\mathrm{\,d}t\\
&=\displaystyle\int\limits_{0}^{1}\dfrac{(t-2)-(t-3)}{(t-2)(t-3)}\mathrm{\,d}t\\
&=\displaystyle\int\limits_{0}^{1}\left(\dfrac{1}{t-3}-\dfrac{1}{t-2}\right)\mathrm{\,d}t\\
&=\left(\ln|t-3|-\ln|t-2|\right)\bigg|_0^1\\
&=2\ln2-\ln3\\
&=\ln4-\ln3=\ln\dfrac{4}{3}.
\end{aligned}$$
Theo đó ta có \(a=1\), \(b=0\), \(c=3\).

Khi đó \(S=1+0+3=4\).