ta có dcos2x=−2sinxcosxdx=−sin2xdx$\rightarrow I=-\int\limits_{0}^{\frac{\pi}2}\frac{dcos^2x}{1+cos^2x}=-(1+cos^2x)|_0^{\frac{\pi}2}=1$
ta có
dcos2x=−2sinxcosxdx=−sin2xdx$\rightarrow I=-\int\limits_{0}^{\frac{\pi}2}\frac{dcos^2x}{1+cos^2x}=-
ln(1+cos^2x)|_0^{\frac{\pi}2}=
ln2$