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Đặt $I=\int\limits_{0}^{\pi/2}\frac{2\sin x - \cos x}{(\cos x + \sin x)^{3}}dx$
Đặt \(t = \frac{\pi }{2} - x\), ta có:
\(I = \int\limits_0^{\pi /2}
{\dfrac{{2\sin x-\cos x dx}}{{{{\left( {\sin x + \cos x} \right)}^3}}} =
\int\limits_{\pi /2}^0 {\dfrac{{2\cos t-\sin t}}{{{{\left( {\sin t + \cos t} \right)}^3}}}\left( { - dt}
\right) = \int\limits_0^{\pi
/2} {\dfrac{{-\sin x+2\cos xdx}}{{{{\left( {\sin x + \cos x} \right)}^3}}}} } }
\)
\( \Rightarrow 2I = \int\limits_0^{\pi /2}
{\dfrac{{dx}}{{{{\left( {\sin x + \cos x} \right)}^2}}}} =\frac{1}2\int\limits_0^{\pi /2} {\dfrac{{dx}}{{c{\rm{o}}{{\rm{s}}^2}\left( {x -
\frac{\pi }{4}} \right)}}\\ = \frac{1}2\tan\left( {x - \frac{\pi
}{4}} \right)} \left| {_0^{\pi /2}} \right. 1\)
\(\Rightarrow I = \frac{1}2\)
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