#int secx/(1+cscx)*dx#
=#int (1/cosx)/(1+1/sinx)*dx#
=#int sinx/[cosx*(1+sinx)]*dx#
=#int (sinx*cosx)/[(cosx)^2*(1+sinx)]*dx#
=#int (sinx*cosx*dx)/((1-(sinx)^2)*(1+sinx))#
=#int (sinx*cosx*dx)/((1-sinx)*(1+sinx)^2)#
After using #y=sinu# and #dy=cosu*du# transforms, this integral became
#int (y*dy)/((1-y)*(1+y)^2)#
Now, I decomposed integrand into basic fractions,
#y/((1+y)*(1-y)^2)=A/(1-y)+B/(1+y)+C/(1+y)^2#
After expanding denominator,
#A*(1+y)^2+B*(1-y^2)+C*(1-y)=y#
Set #x=-1#, #2C=-1#, so #C=-1/2#
Set #x=1#, #4A=1#, so #A=1/4#
Set #x=0#, #A+B+C=0#, so #B=1/4#
Hence,
#int (y*dy)/((1-y)*(1+y)^2)#
=#1/4int dy/(1-y)+1/4int dy/(1+y)-1/2int dy/(1+y)^2#
=#1/4Ln(1+y)-1/4Ln(1-y)+1/2*1/(1+y)+C#
=#1/4Ln((1+y)/(1-y))+1/2*(1+y)^(-1)+C#
Thus,
#int secx/(1+cscx)*dx#
=#1/4Ln((1+sinx)/(1-sinx))+1/2*(1+sinx)^(-1)+C#
