Tính \ I = \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\frac{{\sin \,x\sin \,2xdx}}{{{e^x} + 1}}} .
Giải:
Đặt \ t = - x \Leftrightarrow dx = - dt;x = \frac{\pi }{2} \Rightarrow t = - \frac{\pi }{2};\,x = - \frac{\pi }{2} \Rightarrow t = \frac{\pi }{2}.
\ \Rightarrow I = \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\frac{{\sin \,t\sin \,2tdt}}{{{e^{ - t}} + 1}} = } \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\frac{{{e^t}\sin \,t\sin \,2tdt}}{{\frac{1}{{{e^t}}} + 1}} = } \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\frac{{{e^t}\sin \,t\sin \,2tdt}}{{{e^t} + 1}}} .
\ \Rightarrow 2I = \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\frac{{{e^t}\sin \,t\sin \,2tdt}}{{{e^t} + 1}} + } \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\frac{{\sin \,t\sin \,2tdt}}{{{e^t} + 1}} = } \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\frac{{\left( {{e^x} + 1} \right)\sin \,t\sin \,2tdt}}{{{e^t} + 1}} = \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\sin \,t\sin \,2tdt} } .
\ = \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {2{{\sin }^2}t\,\,d\left( {\sin t} \right)} = \frac{2}{3}{\sin ^3}t\left| \begin{array}{l} \frac{\pi }{2}\\ - \frac{\pi }{2} \end{array} \right. = \frac{4}{3} \Rightarrow I = \frac{2}{3}.
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