$\int\limits_{}^{}\frac{sinxdx}{cosx.\sqrt{2-cos^2x}}$đặt u = cosx $\Rightarrow$ du = -sinxdx
$\int\limits_{}^{}\frac{-du}{u.\sqrt{2-u^2}}=\int\limits_{}^{} \frac{-udu}{u^2.\sqrt{2-u^2}}$
đặt t = $\sqrt{2-u^2} \Rightarrow t^2 = 2-u^2 \Rightarrow tdt = -udu$
$\int\limits_{}^{}\frac{tdt}{(t^2 - 2).t} = \int\limits_{}^{}\frac{dt}{t^2 -(\sqrt{2})^2}= \frac{1}{2.\sqrt{2}}ln\left| {\frac{x-\sqrt{2}}{x+\sqrt{2}}} \right| +C$