Question

sinx分之一的不定积分是多少?

Answer 1

∫1/（sinx）dx

=∫cscxdx

=∫sinx/(1-cos²x) dx

=-∫dcosx/(1-cos²x)

=-1/2[∫dcosx/(1-cosx)+∫dcosx/(1+cosx)]

= -1/2[∫-d(1-cosx)/(1-cosx)+∫d(1+cosx)/(1+cosx)]

=-1/2ln(1+cosx)/ (1-cosx)+C

=ln[(1-cosx)/sinx]+C

=ln(cscx-cotx)+C