integration

\hspace{-16}\int\frac{1}{\sin x+\sec x}dx=\int\frac{\cos x}{\sin x.\cos x+1}dx\\\\\\ \int\frac{2\cos x}{2+\sin 2x}dx=\int\frac{\left(\sin x+\cos x \right)-\left(\sin x-\cos x \right)}{2+\sin 2x}dx$\\\\\\ $\int\frac{\left(\sin x+\cos x \right)}{2+\sin 2x}dx-\frac{\left(\sin x-\cos x \right)}{2+\sin 2x}dx$\\\\\\ $\int\frac{\left(\sin x+\cos x \right)}{3-\left(\cos x-\sin x \right)^2}dx-\int\frac{\left(\sin x-\cos x \right)}{1+\left(\cos x+\sin x \right)^2}dx$\\\\\\ $=I_{1}-I_{2}$ \\\\\\Where $I_{1}=\int\frac{\left(\sin x+\cos x \right)}{3-\left(\cos x-\sin x \right)^2}dx$\\\\\\ and $I_{2}=\int\frac{\left(\sin x-\cos x \right)}{1+\left(\cos x+\sin x \right)^2}dx$

Thank you.

I'll take it from here.