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

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