$I=\dfrac{1}{4} \int_0^{\frac{\pi}{4}} \dfrac{\sin 2x}{\sin 2x + \cos 2x}d(2x) =\dfrac{1}{4}\int _0^{\frac{\pi}{2}} \dfrac{\sin t }{\sin t + \cos t}dt$
Xét $J=\dfrac{1}{4}\int_0^{\frac{\pi}{2}} \dfrac{\cos t}{\sin t + \cos t}dt$
Ta có $I+J = \dfrac{1}{4}\int_0^{\frac{\pi}{2}} dt =\dfrac{1}{4} t \bigg |_0^{\frac{\pi}{2}}=\dfrac{\pi}{8} \ (1)$
$J-I =\dfrac{1}{4}\int_0^{\frac{\pi}{2}} \dfrac{\cos t - \sin t}{\sin t + \cos t}dt =\dfrac{1}{4}\int_0^{\frac{\pi}{2}} \dfrac{d(\sin t + \cos t)}{\sin t + \cos t}dt$
$=\dfrac{1}{4}\ln | \sin t + \cos t| \bigg |_0^{\frac{\pi}{2}}=0 \ (2)$
Lấy $(1)-(2) \Rightarrow 2I = \dfrac{\pi}{8} \Rightarrow I=\dfrac{\pi}{16}$
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