Tính $\ I = \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{4}} {\frac{{x\left( {cos\,4x - cos\,2x} \right) + \sin \,2x}}{{{{\sin }^4}x\left( {2cos2x + 1} \right)}}} dx.$

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Tính $\ I = \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{4}} {\frac{{x\left( {cos\,4x - cos\,2x} \right) + \sin \,2x}}{{{{\sin }^4}x\left( {2cos2x + 1} \right)}}} dx.$
Giải:
Ta có: $\ I = \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{4}} {\frac{{x\left( {2co{s^2}2x - cos\,2x - 1} \right) + \sin \,2x}}{{{{\sin }^4}x\left( {2cos2x + 1} \right)}}} dx = \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{4}} {\frac{{x\left( {2cos2x + 1} \right)\left( {cos2x - 1} \right) + \sin \,2x}}{{{{\sin }^4}x\left( {2cos2x + 1} \right)}}} dx.$


$\  = \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{4}} {\frac{{ - 2x{{\sin }^2}x\left( {2cos2x + 1} \right) + \sin \,2x}}{{{{\sin }^4}x\left( {2cos2x + 1} \right)}}} dx = 2\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{4}} {\frac{x}{{{{\sin }^2}x}}} dx + \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{4}} {\frac{{\sin \,2x}}{{{{\sin }^4}x\left( {2cos2x + 1} \right)}}} dx = 2{I_1} + {I_2}.$

• Xét $\ {I_1} = \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{4}} {\frac{{ - x}}{{{{\sin }^2}x}}} dx.$ Ta đặt: $\ u = x;\,dv =  - \frac{{dx}}{{{{\sin }^2}x}} \Rightarrow du = dx;\,v = \cot \,x.$         

          $$\Rightarrow {I_1} = x\cot x\left| {\begin{array}{*{20}{l}} {\frac{\pi }{4}}\\ {\frac{\pi }{6}} \end{array}} \right. - \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{4}} {\cot {\mkern 1mu} x} dx = \frac{\pi }{4} - \frac{{\pi \sqrt 3 }}{6} - \ln \left| {\sin {\mkern 1mu} x} \right|\left| {\begin{array}{*{20}{l}} {\frac{\pi }{4}}\\ {\frac{\pi }{6}} \end{array}} \right. = \frac{\pi }{4} - \frac{{\pi \sqrt 3 }}{6} - \ln 2$$

• Xét $\ {I_2} = \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{4}} {\frac{{\sin \,2x}}{{{{\sin }^4}x\left( {2cos2x + 1} \right)}}} dx.$Khi đó $\ {I_2} = 2\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{4}} {\frac{{cos\,x}}{{{{\sin }^3}x\left( {3 - 4{{\sin }^2}x} \right)}}} dx.$

         Đặt $\ \,t = \sin \,x \Rightarrow dt = \cos xdx \Rightarrow dx = \frac{{dt}}{{\cos x}}.$

         $\  \Rightarrow {I_2} = 2\int\limits_{\frac{1}{2}}^{\frac{{\sqrt 2 }}{2}} {\frac{{dt}}{{{t^3}\left( {3 - 4{t^2}} \right)}}}  = 2\int\limits_{\frac{1}{2}}^{\frac{{\sqrt 2 }}{2}} {\left[ {\frac{4}{{9t}} + \frac{1}{{3{t^3}}} - \frac{{16}}{{9\left( {4{t^2} - 3} \right)}}} \right]dt} .$

        $$= 2\left( {\frac{4}{9}\ln \left| t \right| - \frac{1}{{6{t^2}}} - \frac{8}{{9\sqrt 3 }}\ln \left| {\frac{{2t - \sqrt 3 }}{{2t + \sqrt 3 }}} \right|} \right)\left| {\begin{array}{*{20}{l}} {\frac{{\sqrt 2 }}{2}}\\ {\frac{1}{2}} \end{array}} \right. = \frac{{2 - 4\ln 2}}{3} + \frac{{16}}{9}\ln \frac{{2 - \sqrt 3 }}{{5 - 2\sqrt 6 }}$$

$\  \Rightarrow I = \frac{{\pi \left( {3 - 2\sqrt 3 } \right)}}{6} + \frac{{2 - 10\ln 2}}{3} + \frac{{16}}{9}\ln \frac{{2 - \sqrt 3 }}{{5 - 2\sqrt 6 }}.$