∫sinx/cosx√(5-4cosx)dx
问题描述：
∫sinx/cosx√(5-4cosx)dx
答
设√(5-4cosx)=t,则sinxdx=tdt/2∴原式=∫(tdt/2)/[t(5-t²)/4]=2∫dt/(5-t²)=(1/√5)∫[1/(√5+t)+1/(√5-t)]dt=(1/√5)[ln│√5+t│-ln│√5-t│]+C (C是积分常数)=(1/√5)ln│(√5+t)/(√5-t)│+C=(1/√...