Question

高数，不定积分，

Answer 1

∫(1+sinx)/[sinx(1+cosx)]dx
＝∫1/[sinx(1+cosx)]dx+ ∫1/(1+cosx)dx
=∫sinx/[sin^2x(1+cosx)]dx+ ∫(1-cosx)/(1-cos^2x)dx
=-∫1/[(1-cos^2x)(1+cosx)]dcosx+ ∫(1/sin^2xdx- ∫1/sin^2xdsinx
=-∫1/[(1-cosx)(1+cosx)^2]dcosx-cotx+1/sinx
=-∫1/[(1-cosx)(1+cosx)^2]dcosx-cotx+1/sinx
=-1/4∫/(1-cosx)dcosx-1/2∫1/(1+cosx)dcosx-1/2∫/[1/(1+cosx)^2dcosx-cotx+1/sinx
=ln(1-cosx)/4-ln(1+cosx)/4+1/2(1+cosx)-cotx+1/sinx+c
=1/4ln(1-cosx)/(1+cosx)+(1+sinx)/2(1+cosx)+c