Đáp án:

$D.\, P = 5$

Giải thích các bước giải:

$\begin{array}{l}\quad I = \displaystyle\int\limits_{\tfrac{\pi}{3}}^{\tfrac{\pi}{2}}\dfrac{\sin x}{\cos2x - \cos x}dx\\ Đặt\,\,u= \cos x\\ \to du = -\sin xdx\\ \text{Đổi cận:}\\ \begin{array}{l|cr} x & \dfrac{\pi}{3}\qquad &\qquad \dfrac{\pi}{2}\\ \hline u&\dfrac12\qquad &\qquad 0 \end{array}\\ \text{Ta được:}\\ I = - \displaystyle\int\limits_{\tfrac12}^0\dfrac{1}{2u^2 - u - 1}du\\ \to I = \displaystyle\int\limits_0^{\tfrac12}\dfrac{1}{2u^2 - u - 1}du\\ \to I = \displaystyle\int\limits_0^{\tfrac12}\dfrac{1}{\left(\sqrt2u - \dfrac{1}{2\sqrt2}\right)^2 - \dfrac98}du\\ Đặt\,\,t = \sqrt2u - \dfrac{1}{2\sqrt2}\\ \to dt = \sqrt2du\\ \text{Đổi cận:}\\ \begin{array}{l|cr} u & 0\qquad &\qquad \dfrac{1}{2}\\ \hline t&-\dfrac{1}{2\sqrt2}\qquad &\qquad \dfrac{1}{2\sqrt2} \end{array}\\ \text{Ta được:}\\ I = \dfrac{1}{\sqrt2}\displaystyle\int\limits_{-\tfrac{1}{2\sqrt2}}^{\tfrac{1}{2\sqrt2}}\dfrac{1}{t^2 - \dfrac98}dt\\ \to I = \dfrac{4\sqrt2}{9}\displaystyle\int\limits_{-\tfrac{1}{2\sqrt2}}^{\tfrac{1}{2\sqrt2}}\dfrac{1}{\dfrac{8t^2}{9} -1}dt\\ Đặt\,\,z = \dfrac{2\sqrt2t}{3}\\ \to dz = \dfrac{2\sqrt2}{3}dt\\ \text{Đổi cận:}\\ \begin{array}{l|cr} t &-\dfrac{1}{2\sqrt2}\qquad &\qquad \dfrac{1}{2\sqrt2}\\ \hline z&-\dfrac{1}{3}\qquad &\qquad \dfrac{1}{3} \end{array}\\ \text{Ta được:}\\ I = \dfrac{2}{3}\displaystyle\int\limits_{-\tfrac{1}{3}}^{\tfrac{1}{3}}\dfrac{1}{z^2 - 1}dz\\ \to I = \dfrac{2}{3}\displaystyle\int\limits_{-\tfrac{1}{3}}^{\tfrac{1}{3}}\dfrac{1}{z^2 - 1}dz\\ \to I = \dfrac{2}{3}\displaystyle\int\limits_{-\tfrac{1}{3}}^{\tfrac{1}{3}}\left[\dfrac{1}{2(z-1)} - \dfrac{1}{2(z+1)} \right]dz\\ \to I = \dfrac{1}{3}\left(\ln|z-1| - \ln|z+1| \right)\Bigg|_{-\tfrac13}^{\tfrac13}\\ \to I = -\dfrac23\ln2\\ \to \begin{cases}a = 2\\b = -3\end{cases}\\ \to a - b = a - (-3)\\ \to P = 5 \end{array}$