Biết I = intlimits_pi/6^pi/3 dx/sin = 1/2(ln a + ln b) tính S=a+b

Đặt  \(t = cosx\Leftrightarrow dt = - \sin {\rm{x}}dx\) và \({\sin ^2}x = 1 - {t^2}.\)  

Đổi cận: \(\left\{ {x = \frac{\pi }{6} \Rightarrow t = \frac{{\sqrt 3 }}{2};x = \frac{\pi }{3} \Rightarrow t = \frac{1}{2}} \right\}\) 

Khi đó \(I = \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}} {\frac{{dx}}{{{\mathop{\rm s}\nolimits} {\rm{inx}}}}} = \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}} {\frac{{\sin x}}{{1 - c{\rm{o}}{{\rm{s}}^2}x}}dx = \int\limits_{\frac{1}{2}}^{\frac{{\sqrt 3 }}{2}} {\frac{1}{{1 - {t^2}}}dt} = \frac{1}{2}.\ln \left. {\left| {\frac{{t + 1}}{{t - 1}}} \right|} \right|} _{\frac{1}{2}}^{\frac{{\sqrt 3 }}{2}}\)

\(= \frac{1}{2}\ln (7 + 4\sqrt 3 ) - \frac{1}{2}\ln 2.\) 

Suy ra \(I = - \frac{1}{2}\left[ {\ln (7 - 4\sqrt 3 ) + \ln 3} \right] = - \frac{1}{2}(\ln a + \ln b)\)

\(\Rightarrow \left\{ \begin{array}{l} a = 7 - 4\sqrt 3 \\ b = 3 \end{array} \right. \Rightarrow a + b = 10 - 4\sqrt 3 .\)