∫[t/(1+cost)]dt=∫[t(1-cost)/sin²t]dt
=∫[t/sin²t]dt-∫[tcost/sin²t]dt
=∫tcsc²tdt-∫[tcost/sin²t]dt
由第一个积分得：
∫tcsc²tdt=-∫td(cott)=-[tcott-∫cottdt]
=-tcott+∫cottdt=-tcott+ln(sint)
由第二个积分得：
∫[tcost/sin²t]dt=-∫td(1/sint)=-t/sint+∫(dt/sint)
=-t/sint+ln|csct-cott|
最后有：
∫[t/(1+cost)]dt=-tcott+ln(sint)+t/sint-ln|csct-cott|+c