题目应该是说求（cosx）^2/（sinx）^2的不定积分.设t=tanx,则x=arctant,dx=dt/(1+t²)∫(cosx）^2/（sinx）^2dx=∫dt/[t²(t²+1)] =∫[1/t²-1/(t²+1)]dt=-1/t-arctant+C,再将t=tanx带回来,得...