Question

用第一类换元积分法求不定积分∫cscxdx,

Answer 1

∫cscxdx＝∫dx/sinx=∫(sinx dx)/sin²x=∫d(cosx)/(cos²x-1)

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

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

=ln√[(1-cosx)/(1+cosx)]+C；