#int sqrt(cos2x)/cosx*dx#
=#int cosx*sqrt(cos2x)/(cosx)^2*dx#
=#int cosx*sqrt[1-2(sinx)^2]/(cosx)^2*dx#
=#int cosx*sqrt[1-2(sinx)^2]/[1-(sinx)^2]*dx#
After using #sqrt2*sinx=sinu#, #sqrt2*cosx*dx=cosu*du# and #sinx=sinu/sqrt(2)# transforms, this integral became
#int ((cosu*du)/sqrt2)*sqrt[1-(sinu)^2]/[1-(sinu/sqrt2)^2]#
#=int (sqrt2*(cosu)^2)/[2-(sinu)^2]*du#
#=sqrt2*int (cosu)^2/[1+(cosu)^2]*du#
After using #z=tanu#, #du=dz/(z^2+1)# and #(cosu)^2=1/(z^2+1)# transforms, it became
#=sqrt2*int (1/(z^2+1))/[1+1/(z^2+1)]*(dz)/(z^2+1)#
=#int (sqrt2*dz)/[(z^2+1)*(z^2+2)]#
=#int (sqrt2*dz)/(z^2+1)-int (sqrt2*dz)/(z^2+2)#
=#sqrt2*arctanz-arctan(z/sqrt2)+C#
=#sqrt2*arctan(tanu)-arctan(tanu/sqrt2)+C#
#=usqrt2-arctan(tanu/sqrt2)+C#
After using #sqrt2*sinx=sinu#, #u=arcsin(sqrt2*sinx)# and #tanu=(sqrt2*sinx)/sqrt[1-2(sinx)^2]=(sqrt2*sinx)/sqrt(cos2x)# inverse transforms, I found
#int sqrt(cos2x)/cosx*dx#
=#sqrt2*arcsin(sqrt2*sinx)-arctan(sinx/sqrt(cos2x))+C#
