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
