#int sqrt(cos2x)/cosx*dx#
=#int sqrt[1-2(sinx)^2]/(cosx)^2*cosx*dx#
=#int sqrt[1-2(sinx)^2]/[1-(sinx)^2]*cosx*dx#
After using #sqrt2sinx=siny# an #sqrt2cosx*dx=cosy*dy# substitution, this integral became
=#int sqrt[1-(siny)^2]/[1-(siny/sqrt2)^2]*((cosy*dy)/sqrt2)#
=#sqrt2int sqrt[(cosy)^2]/[2-(siny)^2]*cosy*dy#
=#sqrt2int ((cosy)^2*dy)/[2-(siny)^2]#
=#sqrt2int (dy)/[2(secy)^2-(tany)^2]#
=#sqrt2int (dy)/[2(tany)^2+2-(tany)^2]#
=#sqrt2int (dy)/[(tany)^2+2]#
=#sqrt2int ([(tany)^2+1]*dy)/([(tany)^2+1][(tany)^2+2])#
After using #z=tany# and #dz=[(tany)^2+1]*dy# transformation, it became
#sqrt2int (dz)/[(z^2+1)*(z^2+2)]#
=#sqrt2int (dz)/(z^2+1)#-#sqrt2int (dz)/(z^2+2)#
=#sqrt2arctanz-arctan(z/sqrt2)+C#
=#sqrt2arctan(tany)-arctan(tany/sqrt2)+C#
=#ysqrt2-arctan(tany/sqrt2)+C#
After using #sqrt2sinx=siny#, #y=arcsin(sqrt2sinx)# and #tany=(sqrt2sinx)/sqrt(cos2x)# inverse transforms, I found
#int sqrt(cos2x)/cosx*dx#
=#sqrt2arcsin(sqrt2sinx)-arctan(sinx/sqrt(cos2x))+C#
