If I=∫(cotx−tanx)dx, then I equals
2log(tanx−cotx)+C
2log|sinx+cosx+sin2x|+C
2log|sinx−cosx+2sinxcosx|+C
2log|sin(x+π/4)+2sinxcosx|+C
I=∫cosx−sinxcosxsinxdx
Put sinx+cosx=t, so that 2sinxcosx=t2−1
∴I=2∫dtt2−1=2logt+t2−1+C=2log|sinx+cosx+sin2x|+C