问题描述: 微积分求原函数的问题求t/(1+cost)的原函数 1个回答 分类:数学 2014-12-06 问题解答: 我来补答 ∫[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 展开全文阅读