Tích phân \(I = }{2}} x}}} \) có giá trị bằng:

* Hướng dẫn giải

Ta có:

\(\begin{array}{l}I = \int\limits_{\dfrac{\pi }{3}}^{\dfrac{\pi }{2}} {\dfrac{{dx}}{{\sin x}}} = \int\limits_{\dfrac{\pi }{3}}^{\dfrac{\pi }{2}} {\dfrac{{\sin x}}{{{{\sin }^2}x}}} \,dx\\ = - \int\limits_{\dfrac{\pi }{3}}^{\dfrac{\pi }{2}} {\dfrac{{d\left( {\cos x} \right)}}{{1 - {{\cos }^2}x}}} \\ = - \dfrac{1}{2}\int\limits_{\dfrac{\pi }{3}}^{\dfrac{\pi }{2}} {\left( {\dfrac{1}{{1 - \cos x}} + \dfrac{1}{{1 + \cos x}}} \right)} \;d\left( {\cos x} \right)\\ = \dfrac{1}{2}\int\limits_{\dfrac{\pi }{3}}^{\dfrac{\pi }{2}} {\dfrac{1}{{1 - \cos x}}d\left( {1 - \cos x} \right)} - \dfrac{1}{2}\int\limits_{\dfrac{\pi }{3}}^{\dfrac{\pi }{2}} {\dfrac{1}{{1 + \cos x}}d\left( {1 + \cos x} \right)} \\ = \dfrac{1}{2}\ln \left| {1 - \cos x} \right|\left| {_{\dfrac{\pi }{3}}^{\dfrac{\pi }{2}}} \right. - \dfrac{1}{2}\ln \left| {1 + \cos x} \right|\left| {_{\dfrac{\pi }{3}}^{\dfrac{\pi }{2}}} \right.\\ = \left( {\dfrac{1}{2}\ln \dfrac{1}{2}} \right) - \dfrac{1}{2}\ln \dfrac{3}{2} = \dfrac{1}{2}\ln \dfrac{1}{3}\end{array}\)