∫cscxdx
=∫[1/(sinx)]dx
=∫1/[2sin(x/2)cos(x/2)] dx
=∫1/[sin(x/2)cos(x/2)] d(x/2)
=∫1/tan(x/2)*sec²(x/2) d(x/2)
=∫1/tan(x/2)d[tan(x/2)]
=ln|tan(x/2)|+C
匿名回答于2020-10-12 01:37:10
