$\ I = \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {\frac{{\sqrt {3\cot \,x + 1} + x}}{{{{\sin }^2}x}}\,dx} .$
Giải:
Ta có: $\ I = \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {\frac{{\sqrt {3\cot \,x + 1} + x}}{{{{\sin }^2}x}}\,dx} = \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {\frac{{\sqrt {3\cot \,x + 1} }}{{{{\sin }^2}x}}\,dx} + \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {\frac{x}{{{{\sin }^2}x}}\,dx} = {I_1} + {I_2}.$
- Xét $\ {I_1} = \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {\frac{{\sqrt {3\cot \,x + 1} }}{{{{\sin }^2}x}}\,dx\,\,} .$
Đặt $\ t = \sqrt {3\cot \,x + 1} \Rightarrow {t^2} = 3\cot \,x + 1 \Leftrightarrow 2tdt = - \frac{3}{{{{\sin }^2}x}}dx \Leftrightarrow dx = - \frac{2}{3}t{\sin ^2}xdt.$
$\ \Rightarrow {I_1} = \frac{2}{3}\int\limits_1^2 {{t^2}\,dt\, = \frac{2}{9}{t^3}\left| \begin{array}{l}
2\\
1
\end{array} \right. = \frac{{14}}{9}\,} .$
- Xét $\ {I_2} = \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {\frac{x}{{{{\sin }^2}x}}\,dx} .$
Đặt $\ \,\,\left\{ \begin{array}{l}
u = x\\
dv = \frac{{dx}}{{{{\sin }^2}x}}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
du = dx\\
v = - \cot \,x
\end{array} \right.$
$\ \, \Rightarrow {I_2} = - x\cot \,x\left| \begin{array}{l}
\frac{\pi }{2}\\
\frac{\pi }{4}
\end{array} \right. + \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {\cot \,x\,dx} \, = \frac{\pi }{4} + \ln \left| {\sin \,x} \right|\left| \begin{array}{l}
\frac{\pi }{2}\\
\frac{\pi }{4}
\end{array} \right. = \frac{\pi }{4} + \ln \sqrt 2 .$
Vậy $\ I = {I_1} + {I_2} = \frac{{14}}{9} + \frac{\pi }{4} + \ln \sqrt 2 .$
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