Finding the indefinite integral of [ln Tan (x / 2)] / SiNx

Finding the indefinite integral of [ln Tan (x / 2)] / SiNx


Let u = ln [Tan (x / 2)], then Du = 1 / SiNx DX
∫ln[tan(x/2)]/sinx dx
=∫u du
=u²/2+C
=½·ln²[tan(x/2)]+C