Đáp án:
$I =\ln\dfrac32$
Giải thích các bước giải:
$\quad I = \displaystyle\int\limits_{\tfrac{\pi}{3}}^{\tfrac{\pi}{2}}\dfrac{1 - \cos x}{\sin x}dx$
$\to I = \displaystyle\int\limits_{\tfrac{\pi}{3}}^{\tfrac{\pi}{2}}\dfrac{1}{\sin x}dx - \displaystyle\int\limits_{\tfrac{\pi}{3}}^{\tfrac{\pi}{2}}\dfrac{\cos x}{\sin x}dx$
$\to I = \ln\left|\tan\dfrac x2\right|\Bigg|_{\tfrac{\pi}{3}}^{\tfrac{\pi}{2}} - \ln|\sin x|\Bigg|_{\tfrac{\pi}{3}}^{\tfrac{\pi}{2}}$
$\to I = \ln\left(\tan\dfrac{\pi}{4}\right)- \ln\left(\tan\dfrac{\pi}{6}\right) - \left[\ln\left(\sin\dfrac{\pi}{2}\right) - \ln\left(\sin\dfrac{\pi}{3}\right)\right]$
$\to I = \ln1 - \ln\dfrac{\sqrt3}{3} - \left(\ln1 - \ln\dfrac{\sqrt3}{2}\right)$
$\to I = \ln\dfrac{\sqrt3}{2} - \ln\dfrac{\sqrt3}{3}$
$\to I =\ln\dfrac32$
