If $\int \frac{\cos x}{\cos 2x}dx=A\ln \left|\frac{1+\sqrt{2}\sin x}{1-\sqrt{2}\sin x}\right|+c,thenA=$

$$\begin{gathered} \mathrm{I}=\int \frac{\mathrm{cosx}}{\cos 2\mathrm{x}}\mathrm{dx}=\mathrm{A} \ln \left|\frac{1+\sqrt{2} Sinx}{1-\sqrt{2} Sinx}\right| +\mathrm{C} \\ \mathrm{I}=\int \frac{\mathrm{cosx}}{\cos 2\mathrm{x}}\mathrm{dx}=\int \frac{\mathrm{cosx}}{1-2 Si{n}^2\mathrm{x}}\mathrm{dx} \\ \mathrm{put} Sin \mathrm{x}=\mathrm{t} \\ \cos \mathrm{x} \mathrm{dx}=\mathrm{dt} \\ \therefore \mathrm{I}=\int \frac{\mathrm{dt}}{1-2{\mathrm{t}}^2}=\int \frac{\mathrm{dt}}{1-{\left(\sqrt{2}\mathrm{t}\right)}^2}=\frac{1}{2\left(1\right)\sqrt{2}} \log \left|\frac{1+\sqrt{2} \mathrm{t}}{1-\sqrt{2} \mathrm{t}}\right|+\mathrm{C} \\ =\frac{1}{2\sqrt{2}}\ln \left|\frac{1+\sqrt{2} Sinx}{1-\sqrt{2} Sinx}\right|+\mathrm{C} \\ \therefore \mathrm{A}=\frac{1}{2\sqrt{2}} \end{gathered}$$