Lời giải:Ta có \(\int\limits_{\frac{\pi }{{4 + {m^2}}}}^{\frac{\pi }{2}} {\left( {2 - \sin 2x} \right){e^{2\cot x}}{\rm{d}}x} = 2\int\limits_{\frac{\pi }{{4 + {m^2}}}}^{\frac{\pi }{2}} {{e^{2\cot x}}{\rm{d}}x} - \int\limits_{\frac{\pi }{{4 + {m^2}}}}^{\frac{\pi }{2}} {\sin 2x{e^{2\cot x}}{\rm{d}}x} .\) (1)
Xét \(\int\limits_{\frac{\pi }{{4 + {m^2}}}}^{\frac{\pi }{2}} {\sin 2x{e^{2\cot x}}{\rm{d}}x} = \int\limits_{\frac{\pi }{{4 + {m^2}}}}^{\frac{\pi }{2}} {{e^{2\cot x}}{\rm{d}}\left( {{{\sin }^2}x} \right)} = {\sin ^2}x.{e^{2\cot x}}\left| {_{\frac{\pi }{{4 + {m^2}}}}^{\frac{\pi }{2}}} \right. - \int\limits_{\frac{\pi }{{4 + {m^2}}}}^{\frac{\pi }{2}} {{{\sin }^2}x\left( { - \frac{2}{{{{\sin }^2}x}}} \right){e^{2\cot x}}{\rm{d}}x} \)
\( = {\sin ^2}x.{e^{2\cot x}}\left| {_{\frac{\pi }{{4 + {m^2}}}}^{\frac{\pi }{2}}} \right. + 2\int\limits_{\frac{\pi }{{4 + {m^2}}}}^{\frac{\pi }{2}} {{e^{2\cot x}}{\rm{d}}x} .\) (2)
Từ (1) và (2) suy ra \(I = {\sin ^2}x.{e^{2\cot x}}\left| {_{\frac{\pi }{{4 + {m^2}}}}^{\frac{\pi }{2}}} \right. = - 1 + {\sin ^2}\frac{\pi }{{4 + {m^2}}}.{e^{2\cot \frac{\pi }{{4 + {m^2}}}}}.\)
\( \to S = \ln \left( {{{\sin }^2}\frac{\pi }{{4 + {m^2}}}.{e^{2\cot \frac{\pi }{{4 + {m^2}}}}}} \right) = 2cot\left( {\frac{\pi }{{4 + {m^2}}}} \right) + 2\ln \left( {\sin \frac{\pi }{{4 + {m^2}}}} \right).\)
