Ta có: $\int\limits_0^{\frac{\pi}{4}}\dfrac{1-2\sin^2x}{1+\sin2x}dx$$=\int\limits_0^{\frac{\pi}{4}}\dfrac{\cos2x}{1+\sin2x}dx$$=\dfrac{1}{2}\int\limits_0^{\frac{\pi}{4}}\dfrac{d(1+\sin2x)}{1+\sin2x}$$=\dfrac{\ln(1+\sin2x)}{2}\left|\begin{array}{l}\dfrac{\pi}{4}\\0\end{array}\right.=\dfrac{\ln2}{2}$
Ta có: $\int\limits_0^{\frac{\pi}{4}}\dfrac{1-2\sin^2x}{1+\sin2x}dx$$=\int\limits_0^{\frac{\pi}{4}}\dfrac{\cos2x}{1+\sin2x}dx$$=\dfrac{1}{2}\int\limits_0^{\frac{\pi}{4}}\dfrac{d(1+\sin2x)}{1+\sin2x}dx$$=\dfrac{\ln(1+\sin2x)}{2}\left|\begin{array}{l}\dfrac{\pi}{4}\\0\end{array}\right.=\dfrac{\ln2}{2}$
Ta có: $\int\limits_0^{\frac{\pi}{4}}\dfrac{1-2\sin^2x}{1+\sin2x}dx$$=\int\limits_0^{\frac{\pi}{4}}\dfrac{\cos2x}{1+\sin2x}dx$$=\dfrac{1}{2}\int\limits_0^{\frac{\pi}{4}}\dfrac{d(1+\sin2x)}{1+\sin2x}$$=\dfrac{\ln(1+\sin2x)}{2}\left|\begin{array}{l}\dfrac{\pi}{4}\\0\end{array}\right.=\dfrac{\ln2}{2}$