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Ta có
$\dfrac{1-\sin^{2}x}{1+\sin2x}=\dfrac{\cos^{2}x}{(\sin x+\cos x)^2}=\dfrac{1}{2}.\left[ {\dfrac{(\sin x+\cos x)\cos x-(\cos x-\sin x)\sin x}{(\sin x+\cos x)^2}+\dfrac{\cos^2 x -\sin^2 x}{(\sin x+\cos x)^2}} \right]$
$=\dfrac{1}{2}.\left[ {\dfrac{(\sin
x+\cos x)(\sin x)'-(\sin
x+\cos x)'\sin x}{(\sin x+\cos x)^2}+\dfrac{(\sin
x+\cos x)'}{\sin x+\cos x}} \right]$
Vậy
$\int\limits_{0}^{\pi/4}\dfrac{1-\sin^{2}x}{1+\sin2x}dx=\dfrac{1}{2}\left[ {\dfrac{\sin
x}{\sin x+\cos x}+\ln \left| {\sin x+\cos x} \right|} \right]_{0}^{\pi/4}=\boxed{\dfrac{1+ \ln 2}{4}}$
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