令u=ln[tan(x/2)],则du=1/sinx dx
∫ln[tan(x/2)]/sinx dx
=∫u du
=u²/2+C
=½·ln²[tan(x/2)]+C弱弱的问一下du=1/sinx dx怎么化的，大神给个过程，谢谢拉du=1/[tan(x/2)]·[tan(x/2)] '=1/[tan(x/2)]·sec²(x/2)·(x/2) '=[cos(x/2)]/[sin(x/2)]·1/cos²(x/2)·1/2 =1/[2sin(x/2)·cos(x/2)]=1/sinx