令t=tan(x/2)，则sinx=2t/(1+t²)，cosx=(1-t²)/(1+t²)dx=2/(1+t²)·dt原式=∫1/[2sinx(1+cosx)]·dx=∫(1+t²)²/(8t)·2/(1+t²)·dt=1/4·∫(1+t²)/t·dt=1/4·ln|t|+1/8·t²+C=1/4·ln|tan(x/2)|+1/8·tan²(x/2)+C
